KINEMATICSExample 3. A particle goes from A to B with a speed of 40 km/h and B to C with a speed of 60 km/h. If AB = 6BC the average speed in km/h between A and C is _______ Sol. AB = 40t1 ...(1) BC = 60t2 ...(2) total distance travelled time taken AB + BC Vav = t1 + t 2 From eqn. (1) and (2) A 40t1 + 60t 2 Vav = ...(3) t1 + t 2 According to question AB = 6BC 40t1 = 6 × 60t2 From eqn (1) and (2) t1 = 9t2 40 9t 2 60t 2 Vav From eqn (3) from eqn (3) 9t 2 t 2 420t 2 Vav = Vav = 42 km / h 10t 2 Average speed = B C Example 6. At a distance L = 400m from the traffic light brakes are applied to a locomotive moving at a velocity v = 54 km/hr. Determine the position of the locomotive relative to the traffic light 1 min after the application of the brakes if its acceleration is –0.3 m/sec2. 5 Sol. u 54 15 m / s 18 a = –0.3 m/s2 v = u + at 0 = 15 – 0.3 t0 15 t0 = = 50 sec 0.3 After 50 second, locomotive comes in rest permanently. v2 = u2 + 2as O2 = 152 – 2 × 0.3 S0 S0 = 225 2250 = = 375m 0.6 6 the distance of the locomotive from traffic light = 400 – 375 = 25 metre Example 7. A car moves in the x–y plane with acceleration (3iˆ + 4 ˆj) m / s 2 . (a) Assuming that the car is at rest at the origin at t = 0, derive expression for the velocity as function of time. (b) Find the equation of path of car and find the position vector as function of time. Sol. Here, ux = 0, uy = 0, uz = 0 2 2 ax = 3 m/s , ay = 4 m/s (a) vx = ux + axt or vx = 3t and vy = uy + ayt or vy = 4t v = v x ˆi + v y ˆj v = (3tiˆ + 4tjˆ) 1 2 a t 2 x 1 3 ´ 3t 2 = t 2 2 2 1 2 uyt + a yt 2 1 (4)t 2 = 2t 2 2 2 4 2´ x = x 3 3 x = uxt + (b) or x= and y= or y= y= 4 x 3 Hence, the path is straight line. The position of car is 3 r = x ˆi + y ˆj = t 2 ˆi + 2t 2 ˆj 2 y= æ 3 2ö ç x= t ÷ è 2 ø The ball is at a height of 80 m at two times. the time interval being 6s. Find u. a = g = – 10 m/s2 and s = 80 m. A disc arranged in a vertical plane has two groves of same length directed 60º along the vertical chord AB and CD as shown in the fig. it is given that u+ u 2 .1600 10 Now. 2 1 SCD = g cos 60º t 2CD 2 A C But SAB = SCD 60 60º º 60º 1 2 gcos g t g AB SAB 2 D = g SCD 1 g cos 60º t 2 CD 2 2 B t AB t AB = 1 : 2 1 = 2 or t CD t 2CD C . Here.1600 10 u 2 .Example 9. A ball is thrown upwards from the ground with an initial speed of u. we have 2 2 80 = ut – 5t or 5t2 – ut + 80 = 0 or t= and t= u+ u- u 2 . The same particles slide D down along AB and CD. Sol. Take g = 10 m/s2. we have u = 50 m/s or A Example 10. 1 Substituting the value in s = ut + at 2 . The ratio of the time tAB/t CD is : (A) 1 : 2 (B) 1 : 2 (C) 2 : 1 (D) 2 : 1 B 1 2 SAB = g t AB Sol.1600 = 30 5 or u2 – 1600 = 900 u2 = 2500 or u = ± 50 m/s Ignoring the negative sign.1600 =6 10 u 2 .1600 =6 or u 2 .1600 u 10 +ve –ve s = 80 m u u 2 . u = u m/s. gt 02 . For second stone.8gh v 0 8gh but they meth at height H in time t0. Find the time when two stones cross each other.. Simultaneously another stone is thrown up from the ground with such a velocity that it can reach a height of 4h.2g (4h) 0 2 = v 02 ... v 2 = v02 ..(1) 2 and that of second stone is 1 H = v0 t 0 .(2) 2 After solving eqn (1) and (2) 1 1 h – v 0 t 0 g t 02 g t 02 or 2 2 h = v0t0 v=0 (1) (2) h v0 4h . Sol. Displacement of 1st stone is 1 h – H gt 20 .Example 11. A stone is dropped from a height h. Sol. t0 = h h = v0 8gh t0 = h 8g Example 12. A rocket is launched at an angle 53º to the horizontal with an initial speed of 100 ms–1.2 152t 0 1 10 t 20 2 1503.2 m O 37º v0 37º h B Engin fail x .h = v 0y t 0 + a y t 02 2 1155. It moves along its initial line of motion with an acceleration of 30 ms–2 for 3 seconds.2 m 10 The maximum altitude reached by the rocket is = h0 + h = (1155. Find : (i) the maximum altitude reached by the rocket (ii) total time of flight. ax = 0 (i) At maximum altitudes vy = 0 2 v 2y = v 0y + 2a y h 0 0 = v 20y + 2a y h 0 h0 – A S0 v20y u 53º 2a y 152 152 h0 2 10 11552 h0 = h 0 = 1155. At this time its engine fails & the rocket proceeds like a free body.2 + 348) m (ii)Total time of flight. 1 y = v0 y t + a y t 2 2 1 . S0 = ut + ½ at 2 = 100 × 3 + ½ × 30 × 9 = 300 + 135 = 435 m In OAB h sin 53º = S0 4 h S0 sin 53º 435 87 4 348 m 5 v0 = u + at = 100 + 30 × 3 = 190 m/s (velocity at the time of switch off) After engin switch off 3 v0x v0 sin 37º 190 114 m / s 5 4 v0y v0 cos37º 190 152 m / s 5 ay = – 10 m/s2. (b) Find the velocity of the particle at origin.2 = 0 t 02 . velocity and acceleration at t = 2 s.. Sol.4 t 0 . (b) the displacement of the particle in the interval t = 5 s to t = 8 s. a = (6 + 18 × 3) cm/s2 a = 60 cm/s2 (b) Given. Find out : (a) the acceleration of the particle at t = 3 s. Sol. x = (2t – 3)2 Velocity.04 = 0 t 0 = 35.1155. (a) Find the position. (a) Position. a = dv = 8 m / s2 dt .1155. v = (3 + 6t + 9t2) cm/s or ds = (3+ 6t + 9t 2 ) dt ds = (3 + 6t + 9t2)dt ò s = éë 3t + 3t 2 + 3t 3 ûù5 or s = 1287 cm or 8 5 ds = 8 ò (3+ 6t + 9t ) dt 2 5 8 Example 14 : The motion of a particle along a straight line is described by the function x = (2t – 3)2 where x is in metres and t is in seconds.231.5t 20 5 t 20 . v dx 4 2t 3 m / s dt and acceleration. (a) Acceleration of particle a= dv = ( 6 + 18t ) cm / s 2 dt At t = 3 s.2 = 152t 0 .30.152 t 0 .54 sec. A particle is moving with a velocity of v = (3 + 6t + 9t2) cm/s. Example 13. x = (2 × 2 – 3)2 = 1.0 m v = 4(2 × 2 – 3) = 4 m/s and a = 8 m/s2 (b) At origin. x = 0 or (2t – 3) = 0 v=4×0=0 .At t = 2 s. A ball is dropped from a height of 80m on a floor. Sol. Sol. 1 1 S 20 25 t 25 v 0 2 2 v0 or 0 250 – t 25 2 500 .2 sec Example 16. From the velocity-time plot shown in figure. Plot the speed-time graph and velocity-time graph of its motion till two collisions with the floor.. .. t = 36. 1 S1 v t 2 1 S1 5 20 S1 = 50 m 2 The distance travelled by the partile during next 20 second is 1 S2 5 20 2 S2 = 50 m Since distance is a sclar quantity therefore total distance = S1 + S2 = 50 + 50 V 5m/s –5m/s S = 100 m Example 17.. Find the time when particle returns to the starting point. Since.(1) or t – 25 v 0 v 20 0 Also tan from figure 25 20 t 25 v0 or 4 t – 25 .(2) v 0 4 t – 25 From eqn (1) and (2) 500 125 t – 25 4 t – 25 t – 25 10 10 20 25 t v 20 20 25 t 20 t(s) 40 v0 t – 25 2 125 t – 25 125 5 5 t = (25 + 5 5 ) sec. Also find the average velocity during this period. At each collision the ball losses half of its speed. total displacement is zero. The fig. [Take g = 10 ms–2].v 20 Example 15. The distance fravelled by the particle during the first 20 second. find the distance travelled by the particle during the first 40 seconds. the area of v – t graph gives displacement. When the particle comes at initial position. shows the v–t graph of a particle moving in straight line.. In this case area of v – t graph should be zero. it is dropped again from 20m with zero initial speed. It now loses half of initial speed after the collision i. Time taken in reaching the ground 1 20 = 0 + ´ 10(t) 2 2 t=2s Also final speed v´2 = 0 + 2 × 10 × 20 (from v2 = u2 + 2gh) v´ = 20 m/s Thus. Calculate the distance covered by the object : 20 ms–1 A B A´ 2 5 v O 0 B´ C 10 .´ 10´ (2) 2 2 h´ = 40 – 20 = 20 m Now. The velocity-time graph of an object moving along a straight line is as shown in the fig. [Take downward motion positive and upward motion negative in case of v–t graph] Velocity (m/s) 40 20 4 6 Time(s) 8 –20 Example 18. when it first bounces its initial speed is 20çè = 2ø So. we can now draw the velocity-time graph. we can draw the speed-time graph..Sol. Speed (m/s) 40 20 4 6 Time(s) 8 Since. with the above data. velocity is a vector quantity so from the above graph.e. it attains height 1 h´= 20´ 2 . the time is loosing half of its speed. The time in first collision 1 2 gt (during downward motion) 2 1 80 = 0 + ´ 10 ´ t 2 ( u = 0) or 2 2´ 80 t= = 4s or 10 Final speed just before first collision v = 0 + 10 × 4 = 40 m/s h = ut + æ 40 ö ÷m / s. 0 = 20 – 10 × t´ (during upward motion) 20 t´= = 2s (final speed = 0) 10 In 2 s. (a) Let x1 be the distance covered in the time interval between t = 0 to t = 5 s. Then. AB + OC 3 + 10 = ´ AA´= ´ 20 = 130 m 2 2 . Then. Sol. x2 = area of the trapezium OABC. x1 = area of the trapezium OABB´ æ AB + OB´ ö 3+ 5 =ç ´ 20 = 80 m ÷´ AA´= è ø 2 2 (b) Let x2 be the distance covered in the time interval between t = 0 to t = 10 s.(a) between t = 0 to t = 5 s and (b) between t = 0 to t = 10 s. . . A’s speed is 10m/sec in a direction PQ . The velocity of C relative to B is 12 m/sec in a direction QP . A.v B 55 351 ˆ v C = vCB + v B = .vA or 6cos i 6sin j v B 10 ˆi v B cos iˆ 6sin ˆj 10 ˆi 6 = 15 ˆ 351 ˆ i6 j 10 ˆi 24 24 351 ˆ 15 ˆ 4 10 i 4 j 55 ˆ 351 ˆ i j 4 4 v CB = vC . v A = 10 ˆi v B 6 cos ˆi 6sin ˆj 15 351 Here cos .12 iˆ + ˆi + j 4 4 7 351 ˆ = ˆi + j 4 4 But 2 7 351 vC 4 4 = 49 + 351 = 4 2 400 20 = 4 4 Ans. Sol.Example 21. then find the magnitude of the velocity of C. The velocity of B relative to A is 6 m/sec at an angle of cos–1(15/24) to PQ. 5 m/s . sin 24 24 And v CB = . B & C are three objects each moving wit constant velocity.12 ˆi v BA = v B . Example 23. it would take to that if he would have denied the passengers. starts perpendicular to flow of river 200m wide and flowing with 2m/s. Actual velocity of boat should be along BC. AD = vx t = 2t = 50 m Event (2) — From B to C. vy 4 Also. A man with some passengers in his boat. (a) Event (1) — From A to B. Sol. v rel = v b . boat speed in still water is 4m/s. This actual velocity is found by resultant of vrel and vr. (a) Find the direction due to which he must row to reach the required end. But . Time taken by boat to recent from A to B is 100 100 t= = = 25sec.vr x–comput of actual velocity is y vx = vr – vrel sin = 2 – 4sin C E vb and y – comput of actual velocity is B 200m vrel vy = vrel cos vrel vr The time taken to go from B to C is D A BE t0 = vy 100 100 v rel cos 4 cos 25 cos EC = –vx t 0 = –(2 – 4 sin ) t 0 25 4 sin 2 cos EC = AD = 50 m Also. When he reaches half the width of river the passengers asked him they want to reach the just opposite end from where they have started. (b) How many times more total time. Let v M = velocity of man = 6 km/h v = relative velocity of rain w. 4 If boat cross the river with final condition.r. Find the actual direction of rain.1= 5 5 25 125 200 t 2 = 25 + = 25 + = 3 3 3 5 t2 4 4 t2 200 = t1 3 t1 3 50 3 t1 = Example 24. To a person going west wards with a speed of 6 km/h rain appears to fall vertically downwards with a speed of 8 km/h. t 2 = t AB + t BC t2 = t + t0 25 t 2 25 cos 1 tan 2 2 2 cos 2 5 cos 2 cos 2 1 2 4 2 1 5 8 3 = . Sol.t. man = 8 km/h . (b) 25 cos 25 50 4 sin 2 cos 50 cos = 100 sin – 50 cos = 2 sin – 1 1 + cos = 2 sin 2 cos 2 4sin cos 2 2 2 a 1 tan = 2 2 1 tan 1 2 2 EC 4 sin 2 1 2 tan 1 2 If boat crosses the river with initial condition 200 200 t1 = = vy vrel 200 = 50sec. v a S v r = . Sol. The pilot wishes to travel from A to a point B north east of A. vaw = 400 km/h and v a should be along AB or in north-east direction.20kˆ +15iˆ .20 kˆ z N v m = 5 ˆi E v a = 15 ˆi x W v ra = v r . Given that vw = 200 2 km / h . the resultant of vectors OA and OB as shown in figure. v ra = .20 kˆ + 10 ˆi 10 1 tan 20 2 1 tan1 2 vrm vertically Example 26.20 kˆ + 15 ˆi v rm = v r .5iˆ = .Vertical v R = actual velocity of rain N In this case v = v R + (. Sol. Find the direction he must steer and time of his journey if AB = 1000 km. A wind of 200 2 km / h is blowing from the south. N B va A 45º 45º vw = 200 2 km/h C vaw = 400 km/h E . the direction of v aw should be such as the resultant of v w and v aw is along AB or in north-east direction.v M ) vM O W E A v = vR .v m = . A person is running in the rain with a velocity of 5 ms–1 and a wind is also blowing with a speed of 15 ms–1 (both towards east). If is the angle that v R makes with the vertical.75 OB | v | 8 or = 36º 52´ (east of vertical) Example 25. An aircraft flies at 400 km/h in still air.vM v vR B or C vR = vM + v S S v R = (6) 2 + (8)2 = 10 km / h The velocity of rain ( v R ) is given by the vector OC . Rain is falling vertically with a speed of 20 ms–1 relative to air. Thus. then BC | v M | 6 tan 0. Find the angle with the vertical at which the person should hold his umbrella so that he may not get drenched. | va | 400 Further. Applying sine law in triangle ABC.83 h | va | 546.67 Example 27.707 h h km | va | 546. do not enter the cart ? Sol.Let v aw makes an angle with AB as shown in fig. The cart moves uniformly along the horizontal path with a speed of 6 m/s. v c = 6 ˆi v r = 2 ˆj v rc = v r .vc vrc v 6 tan c 3 vr 2 AE AE cos ED sin tan BE cos sin 3 2sin 2 / 2 2sin cos 2 2 cot 3 2 2sin 2 2 v=6m/s vc D x vr y cos A cos sin E vr B C D . A glass wind screen whose inclination with the vertical can be changed. sin(180º 45º 30º ) sin 45º sin105º km | v a | (400) or sin 45º h km 0. is mounted on a cart as shown in figure.47 h The time of journey from A to B is AB 1000 t h t = 1. At what maximum angle to the vertical can the wind screen be placed so that the rain drops falling vertically downwards with velocity 2 m/s.9659 km cos15º | va | (400) (400) sin 45º 0. we get AC BC sin 45º sin BC sin sin 45º or AC 200 2 1 1 sin 400 2 2 = 30º therefore. the pilot should steer in a direction at an angle of (45º + ) or 75º from north towards east. or or or 3 2 1 tan 2 3 1 tan 1 2 3 cot 1 2 tan 1 3 . We have given v = t 2 ˆi + 3t ˆj Here. Since. Sol. vy = 3t and vz = 0 vx = t2 or dx t2 dt or 0 dx 0 t dt or Also. vx = t2. Example 28. A bird flies in the x–y plane with a velocity v = t 2 ˆi + 3t ˆj . x t 2 t3 3 vy = 3t x or dy 3t dt or 0 dy 0 3t dt y t 3t 2 2 Thus. Calculate position and acceleration of bird as function of time. position of bird is r x ˆi y ˆj t3 3t 2 ˆ r ˆi j 3 2 vx = t2 y and dv x d(t) 2 2t dt dt vy = 3t ax . bird is at origin. At t = 0. acceleration of bird is a a x ˆi a y ˆj a 2t iˆ 3 ˆj or 3 .dv y dt dt dt ay = 3 unit Thus. Example 31. 1 2 2g 2g Clearly. Prove that the maximum horizontal range is four times the maximum height attained by the projectile. A particle is projected upwards with a velocity of 100 m/sec at an angle of 60º with the vertical. the horizontal range is maximum and is given by R max u2 g Maximum height attained or u 2 sin 2 45º u 2 R max H max 2g 4g 4 Rmax = 4 Hmax Example 30.Example 29. taking g = 10 m/sec2. Find the time when the particle will move perpendicular to its initial direction. Sol. Here ax = 0 ay = –g ux = 100 sin60º = 50 3 uy = 100 cos60º = 50 . There are two angles of projection and 90º – for which the horizontal range R is same. Show that the sum of the maximum heights for these two angles is independent of the angle of projection. the sum of the heights for the two angles of projection is independent of the angle of projection. There are two angles of projection for which the horizontal range is the same. H1 u 2 sin 2 2g u 2 sin 2 90º H2 and 2g 2 2 u cos H2 2g 2 u u2 2 2 H H sin cos Therefore. when fired at an inclination so as to have maximum horizontal range. For = 45º. Sol. Now. Sol. Thus.8 560 20 sin 1 (82)2 20 = sin–1(0.(ii) 1 0 (46. (i) with . the horizontal displacement is the range..(i) which gives us gR 20 sin 1 2 v0 9.. u v0 or or 50 60º x 3 ˆi 50 ˆj 50 3 ˆi 50 gt ˆj 0 7500 + 2500 – 500 t = 0 10000 t= 500 t 0 = 20 second Example 32.7º ) 23º 2 and = 90º – 0 = 90º – 23º = 67º The commandant of the fort can elevate the cannon to either of these two angles and (if only there were no intervening air!) hit the pirate ship. (b) We have seen that maximum range corresponds to an elevation angle 0 of 45º. shows a pirate ship 560 m from a fort defending the harbor entrance of an island. fires balls at initial speed v0 = 82 m/s. A defense cannon. from Eq. (a) Because the cannon and the ship are at the same height. The horizontal range is R v 20 sin 2 g . Fig.gt )ˆj But u and v are perpendicular.816) . (a) At what angle 0 from the horizontal must a ball be fired to hit the ship ? (b) How far should the pirate ship be from the cannon if it is to be beyond the maximum range of the cannonballs? y 6 3º 27º x R = 560 m Sol.. located at sea level.. u 0 = u x ˆi + u y ˆj = 50 3 ˆi + 50ˆj v y = u y + a y t = 50 – gt y v x = u x = 50 3 m / s v = 50 3 ˆi + (50 .. R .8 R = 686 m 690 m As the pirate ship sails away.0 = 45º. the two elevation angles at which the ship can be hit draw together. Beyond that distance the ship is safe. eventually merging at 0 = 45º when the ship is 690 m away. v02 (82)2 sin 20 sin(2 45º ) g 9. Example 33. it moves horizontally.. A particle is projected in the X–Y plane. 4 sec after projection. So that. 2 sec after projection the velocity of the particle makes an angle 45º with the X-axis. Sol.(1) 2 u cos = v cos 45º v sin45 = 45 sin – 10 × 2 . Find the velocity of projection.. T y =4 2 v0 2u sin 2u sin T8 T g g O x 80 u sin 40 . After 4 sec the particle reach at maximum hight. At maximum height it move horizontally. ..u sin 20 20 u cos u cos From eqn (1) and (2) u2 sin2 + u2 cos2 = 402 + 202 u2 = 1600 + 400 = 2000 1 u 20 100 10 20 20 5 m / s ucos = 20 .(2) . . Find . A train starts from rest and moves with a constant acceleration of 2.Example 34. the projectile hits the ground with a velocity 98 2 m / s at an angle of = 45º with horizontal. Example 35.8 m/s2) Sol. applying 1 sy u y t a y t2 2 1 490 0 (9. sy = 490 m. uy = 0 and ay = g (a) At A. Here.0 m/s2 for half a minute. Find (a) the time taken by the projectile to reach the ground (b) the distance of the point where the particle hits the ground from foot of the hill and (c) the velocity with which the projectile hits the ground (g = 9.8)t 2 2 t = 10 s 1 BA s x u x t a x t 2 (b) 2 or BA = (98) (10) + (0) or BA = 980 m (c) vx = ux = 98 m/s vy = uy + ayt = 0 + (9. The brakes are then applied and the train comes to rest in one minute.8) (10) = 98 m/s v v x2 v 2y (98)2 (98)2 98 2 m / s and tan = 45º vy vx O u = 98 m/s x y B A vx vy 98 1 98 Thus. ux = 98 m/s. A projectile is fired horizontal with a velocity of 98 m/s from the top of a hill 490 m high. So. ax = 0. Sol. v22 = u2 + 2aS (20)2 = 0 + 2 × 5 × S 400 S= S = 40 m 10 Car maintains a constant velocity of 20 m/s for 5 Second.7 km (b) the maximum speed attained by the train is u 2 = 60 m / s (c) the position of the train at half of minimum speed. Draw a velocity-time graph and find the distance covered by the car. Sol. S’ = v × t S '' = 100 m S’’ = 20 × 5 . 1 S2 = u2t2 + a2t22 2 u2 = u1 + a1t1 u2 = 0 + 2 × 30 u2 = 60 m/s v2 = u2 – a2t2 0 = 60 – a2 × 60 a 2 = 1 m / sec2 v22 = u22 + 2 a2S2 0 = 60 × 60 – 2 × 1 × S2 S2 = 1800 m Total distance(s) = S1 + S2 S = (900 + 1800) m S = 2700 m S = 2. It maintains a constant velocity of 20 ms–2 for 5 seconds and then is brought to rest again by a constant acceleration of –2 ms–2. (a) S1 = The distance moved by the train in first half minute. 1 S1 = u1t1 + a1t12 2 1 1 S1 = 0 × t1 + × 2 × (30)2 { t1 = minute} 2 2 1 S1 = × 2 × 900 S1 = 900 m 2 The distance moved by the train after brakes applied. (b) the maximum speed attained by the train and (c) the position(s) of the train at half the maximum speed.(a) the total distance moved by the train. A car is moving along a straight line. It is taken from rest to a velocity of 20 ms–1 by a constant acceleration of 5 ms–2. u22 = u12 + 2a1S’ (30)2 = 0 + 2 × 2 × S’ 900 = S' S' = 225 m 4 Example 36. = S + S’ + S’’ = 40 + 100 + 100 = 240 m Example 37. 2 1 2 h AB gt AC 2 2h AB t AC g Further.0 v 5.9 s 9. Sol.55 m / s or t AC 0. ay= –g – 50 vy 2 But – tan 45º v x v 0 cos 30 v v 0 cos 30 0 50 2 1 3 v0 50 2 2 v0 = 100 3 +1 100 = 3 –1 3 1 50 3 –1 3 –1 . Using h or 1 2 gt we have.The car comes to rest by a constant acceleration of –2 m/s (v’)2 = v2 + 2 × a’S’’ 0 = (20)2 – 2 × 2 S’’ 400 S '' = S'' = 100 m 4 The total distance covered by the car. If it strikes the floor at a point 5m horizontally away from the edge of the table.8 BC 5. A ball rolls off the edge of a horizontal table top 4m high. 24 0. what was its speed at the instant it left the table ? Sol.9 BC = vtAC v A 4m C B 5m Example 38. A ball is projected at an angle of 30º above with the horizontal from the top of a tower and strikes the ground in 5 sec at an angle of 45º with the horizontal. Find the height of the tower and the speed with which it was projected. ux = v0 cos30º uy = v0 sin30º Vx = uX = v0 cos30º y vy = u y + a y t v0 vy = v0 sin30 – gt 30º x v0 H v y 10 5 2 v0 ax=0. the two projectiles are fired simultaneously. vx = v0cos37º vy = v0sin37º y ax = gsin37º = 6 m/s2 v0 O 37º 2 ay = –gcos37º = –8 m/s From O to A. A ball is thrown horizontally from a cliff such that it strikes ground after 5 sec. vertical component of velocities of both projectiles should be same . When two projectiles are projected from same height.} 0 = uy × 5 + ½ × –8 × 25 5uy = ½ × 8 × 25 100 uy = = 20 m / s 5 3 20 v0 5 100 v0 = m/s 3 37º A x Example 40. 1 y = . 20m/s What should be the initial speed of the left side projectile for the two 60º 45º projectile to hit in mid-air ? 10 m Sol. A ball is projected from top of a tower with a velocity of 5 m/s at an angle of 53º to horizontal. What is the initial velocity of projection. 37º y = uyt + ½ ayt 2 { t = 5 sec. According to conservation principle of machenic energy Ui + T i = Uf + T f 1 1 0 + mu 2 = mgh + mv 2 2 2 2 2 u = 2gh + v v = u 2 . Sol. displacement along y-axis is zero.45 m from the point of projection is : (A) 2 m/s (B) 3 m/s (C) 4 m/s (D) data insufficient Sol.Also. Its speed when it is at a height of 0. In the figure shown. then for collision.45 v 25 9 16 v 4 m/s u Example 41.H = u yt + a yt2 2 1 H v0 sin 30 5 10 52 2 H 125 2 – 3 m Example 39.2gh v 52 2 10 0. The line of sight from the point of projection to the point of hitting makes an angle of 37º with the horizontal. (2) 2 2 u cos v 2x v 2y 5 2 H u cos u 2 cos 2 u 2 sin 2 2g 5 2 u cos 2 5 2 u 5 u2 g 1 u 2 sin 2 2g sin 2 2 2 sin 1 5 2 2 2 sin cos2 1 5 2 cos 2 sin 2 or . (1) and (2)} sin 2 sin 2 tan cos tan u 2 sin 2 g tan sin 2 2 tan cos sin 2 2 5 5 2 tan tan 60º 2 2 . the angle of projection is _____ and the velocity vector angle at half the maximum height is ______.. 5 5 or 5 5sin 2 sin 2 2 or 3 – 4 sin2 = 0 3 sin 60º 2 vy u 2 sin 2 gH tan and vx u cos cos 2 u 2 sin 2 2g u cos 1 sin 2 {from eqn. The speed of a particle when it is at its greatest height is 2 / 5 times of its speed when it is at its half the maximum height. uy = usin u u cos ux = ucos vx = ucos .(1) 2 2 vy = uy +2ayS H v 2y u 2 sin 2 2g .. Sol.. u sin 60º = 20 sin45º 20sin 45º u= sin 60º 20 2 u 2 3 u = 20 2 m/s 3 Example 42.. . relative to the elephant.8 180 v 20 9 v0 = 3 m / s or 136 Example 44.8 8 In BAC 1 1. Sol.8 20 or 60 3600 2 72 64 2 24 2 64v 0 v0 + = 6. The line of greatest slope of incline lies in the plane of motion of projectile.8 0 10 t 3 53º 37º x 53º 6 60 grad B t= t v0 = . A dear running with a speed V in front at a distance of 4 5m moving perpendicular to the direction of motion of the elephant.7 .6 6 3 1 2 vy = u y + a y t gt 2 C 0. A hunter is riding an elephant of height 4m moving in straight line with uniform speed of 2 m/ sec.8 or v 0 6. Calculate the initial velocity of projection if its hits perpendicularly an incline of angle 37º which starts from the ground at the bottom of the wall.7 .8 180 60 180 6.. then at what angle to it’s direction of motion must he thrown his spear horizontally for a successful hit.. Find also the speed ‘V’ of the dear.gt 2 1.8 – 20t 8v 64 v02 3v 0 0 6..tan 3 2 tan 1 3 2 Example 43.8v1 = 10 t v1 h = 1. A projectile is to be thrown horizontally from the top of a wall of height 1.(1) v0 O x 0.8 v1 vx = u x = v 0 or 0.5t 2 AB 3 2 tan 37º = = = AC 4 uxt v0 t 2 or 3v0t = 6. ux = v0 uy = 0 ay = g = 10 m/s2 vx = v1 cos53º = 0. h = gt 2 2h 24 8 t g 10 10 . If hunter can throw his spear with a speed of 10 m/sec.7 m.6 v1 = v0 v 10v0 5v 0 v1 = 0 = = .6 v1 vy = v1 sin53º = 0. 1 2 Sol.7 m A 37º 5v 0.(2) y v1 0. . displacement of deer and displacement of spear along y–axis will be same in time t.. So. along x–axis : ux t = 4 5 8 4 5 10 10 4 5 10 10cos 2 4 5 or 2 8 10 cos = 8 8 4 cos = 37º 10 5 From eqn (1) or 10cos 2 v = 10 sin = 10 sin37º = 10 3 = 5 4m 2m/s x 6 m/s y A Example 45. All the motions are in x – y plane.. At the same instant a stone is x O 3. v t = u yt v = 10 sin . Take g = 10 m/s2.Assume horizontal plane at x–y–plane.25 If the stone hitt the object after time t. Find u and the time after which the stone hits the object.25 (u sin )t gt 2 1 1.25 (u sin )t 10 t 2 2 1.25 m on a P 1.25 + 5t 2 . ux = u cos uy = u sin ax = 0 u u =ucos 45º ay = –g 1. So that virtical displacement of stone is 1.(1) Also.5 m/s2.25 m u cos 1 2 y = uyt + a yt 2 1 2 therefore 1.25 = (u sin) t – 5 t 2 (u sin)t = 1.. v rel 10 cos ˆi 10sin ˆj v rel u 2 ˆi u v rel 2 ˆi 10 cos 2 iˆ 10 sin ˆj The deer is moving along y–axis. Sol.(1) Hotizontal displacement of stone is x = 3 + displacement of object A. Initial velocity of object is zero. A stationary person on the ground observes the stone hitting the object during its downward motion at an angle of 45º to the horizontal.0m projected from the origin with a velocity u as shown. An object A is kept fixed at the point x = 3 m and y = 1.25m plank P raised above the ground. At time t = 0 the plank starts moving along u the + x direction with an acceleration 1. u sin x . 25 t2 = 1 From eqn (1) and (2) uy = u sin = 6.25) 2 + (3..75 t 2 (u cos)t = 3 + 0..25 + 5..(2) .75 t 2 Since velocity vector inclined at 45º with horizontals.(4) t = 1 sec.25 m/s ux = 3.75) 2 u = 7. u y u sin gt tan(–45º ) ux u cos u cos = – (u sin – gt) u cos = gt – u sin (u cos) t + (u sin) t = 10 t2 Add eqn (1) and (2) (u cos) t + (u sin) t = 4.29 m / s ..5 t 2 0.75 m/s .so displacement of object is 1 1 1.75 t2 from eqn (3) and (4) 10 t2 = 4.25 + 5 .75 t 2 2 2 x = 3 + 0.27 t2 4.25 m/s uy = 6.25 t2 = 4. u = u 2x + u 2y u = (6.(3) ..25 at 2 1.. . . Sol. which is also moving with acceleration 10 m/s2 vertically upward. u sin = 20 sin 30º = 10 m/s Example 48. The particle hits the inclined plane at an angle of 30º. ux = 10 cos30º = 5 3 m/s uy = 10 sin30º = 5 m/s arel =20 m/s2 10m/s2 30º .76 s 9..8 Example 47. A particle is projected up an inclined plane with initial speed v = 20 m/s at an angle = 30º with plane.3 m 2 9. A particle is thrown horizontally with relative velocity 10 m/s from an inclined plane. Find the time after which it lands on the plane (g = 10 m/s2). The particle hits the plane at 30º (the angle of inclination of plane). It means particle hits the plane horizontally.8 u 2 sin 2 H 2g t (b) 30º 30º 2 20 sin 2 60º H 15. Component of velocity perpendicular to plane remains the same (in opposite direction) i. during its journey. Sol. Sol. (b) the height of the point of impact from the horizontal plane passing through the point of projection. Find the component of its velocity perpendicular to plane when it strikes the plane.e.Example 46. A particle is projected with a velocity of 20 m/s at an angle of 30º to an inclined plane of inclination 30º to the horizontal. Find the (a) time of impact. (a) t u T u sin 2 g 20sin 30º 30º 1. y g 45º 45º 5 2 5 2 2 1 sec. 10 m / s 8 m/s º 37 . vy = uy + ayt vy = 0 + 10 sin37º × 1 3 vy = 10 1 7º 5 in3 gs vy = 6 m/s 37º vx = ux = 8 m/s x y v = v x2 + v 2y = 82 + 6 2 Ans. A ball is projected on smoot inclined plane in direction perpendicular to line of greatest slope with velocity of 8m/s.y 10sin30º 10 m/s2 v0= 10m/s KINEMATICS O 30º 0º o s3 c A 10 ax = arel sin30º = 10 m/s2 ay = – arel cos30º 3 10 3 m / s 2 2 when ball lands on inclined plane. y = 0 1 y = uyt + ay t 2 2 1 0 5 t 10 3 t 2 2 5 3 t =5 1 t= sec. Find the time of the flight of the particle. u0x = 5 2m / s u0y = 0 ax = –g sin45º ay = –g cos45º At point Q. vx = 0 { the particle collied normally at point Q} The time taken by particle to go from P to Q is t0. 3 20 x 30º Example 49. It collides at Q normally. Sol. A particle is projected from point P with velocity 5 2 m/s perpendicular to the surface of a hollow right angle cone whose axis is vertical. Find it’s speed after 1 sec. g sin 45º 10 t 0 = 1 sec. t0 Example 50. vx = u x + a x t 0 = 5 2 – g sin45º t 0 P Q 45º y x x P Q 45º Sol. . . . . Assertion (A) : Reason (R) : A body can have acceleration even if its velocity is zero at a given instant of time. Rocket takes flight due to combustion of fuel and does not move under the gravity effect alone. (A) (B) (C) (D) (E) 4. v 2 u 2 2as . then velocity at displacement s comes out to be. (A) (B) (C) (D) (E) 6. (A) (B) (C) (D) (E) . Assertion (A) : Reason (R) : Displacement of a body may be zero when distance travelled by it is not zero. 2 v S0 Reason (R) : S v 2 versus s graph is a straight line passing through origin with positive intercept and negative slope. If we draw a graph between v 2 and s. Reason (R) : For maximisation of range sin 2 should be equal to one. Assertion (A) : Horizontal component of the velocity of angular projectile remains unchange during its flight. At highest point the velocity reduces to zero. If A is true but R is false. (A) (B) (C) (D) (E) 3. it will be a straight line as shown in figure. (A) (B) (C) (D) (E) Reason (R) : 7. its range is maximum. Assertion (A) : Reason (R) : Rocket in flight is not an illustration of projectile. (A) (B) (C) (D) (E) 5. THE NEXT QUESTIONS REFER TO THE FOLLOWING INSTRUCTIONS A statement of assertion (A) is given and a Corresponding statement of reason (R) is given just below the statements. A body is momentarily at rest when it reverses its direction of motion. If A is false but R is true. The displacement is the longest distance between initial and final position. Slope of displacement-time graph = Velocity of the body. If both A and R are true but R is not the correct explanation of A. If both A and R are false. Assertion (A) : Reason (R) : 2. (A) (B) (C) (D) (E) Suppose a particle starts moving in a straight line with initial velocity +u and an acceleration –a. Assertion (A) : When a body is projected with an angle 45°. Assertion (A) : The slope of displacement-time graph of a body moving with high velocity is steeper than the slope of displacement-time graph of a body with low velocity. mark the correct answer as – If both A and R are true and R is the correct explanation of A.ASSERTION & REASON it of (A) (B) (C) (D) (E) 1. The displacement is proportional to time for uniformly accelerated motion. Assertion (A) : Reason (R) : The speedometer of an automobile measure the average speed of the automobile. Assertion (A) : Reason (R) : A body falling freely may do so with constant velocity. The body falls freely. (A) (B) (C) (D) (E) Assertion (A) : Reason (R) : The displacement-time graph of a body moving with uniform acceleration is a straight line. (A) (B) (C) (D) (E) . Assertion (A) : Reason (R) : 9. Sometimes relative velocity between two bodies is equal to difference in velocities of the two. The relative velocity between any two bodies moving in opposite direction is equal to sum of the velocities of two bodies. when acceleration of a body is equal to acceleration due to gravity. (A) (B) (C) (D) (E) 10. (A) (B) (C) (D) (E) 11. Average velocity is equal to total displacement per total time taken.8. 9 s (C) 9 s (D) 10 s 5.Level # 1 1. (C) The distance traveled by a freely falling stone released. the total no. The train travels with constant velocity for . With zero velocity in the last second of its motion to that traveled by it in the last second of its motion to that traveled by it in the last but one second is 7 : 5. It is at distances b. A train starting from rest travels the first part of its journey with constant acceleration a. Two cars A and B. (D) None of these . The stone strike the ground with velocity 39. then which of the following relations is true ? (A) a(y – z) + b(z – x) + c(x – y) = 0 (B) a(x – y) + b(y – z) + c(z – x) = 0 (C) a(z – x) + b(x – y) + c(y – z) = 0 (D) ax + by + cz = 0 4. The average speed for 7v . the bird directly flies back to A and so forth.. each having a speed of 30 km/hr are heading towards each other along a straight path. the bird directly flies back to A and so forth. A body of mass 3 kg falls from the multi-storeyed building 100 m high and buries itself 2m deep in the sand. v1 v 2 2 v1 v 2 3 v1 v 2 (C) (D) Zero (B) (A) v1 v 2 v1 v 2 v1 v 2 8. The time of penetration will be (A) 0. being brought to rest. 2n. 3n second. of trips which the bird makes till the cars meet is (A) Four (B) Eight (C) Sixteen (D) Infinite 7. heads directly towards car B. yth and zth second from start. Three particles start from the origin at the same time. one with a velocity v 1 along x-axis the second along the y-axis with a velocity v 2 and the third along x = y line. of the total time the whole journey is 8 (A) 2. d from the same point after n. The velocity of the third so that the three may always lie on the same line. on reaching B. 2 (B) A ball falls from the top of a tower in 8 second in 4 second. A bird that can fly at 60 km/hr flies off car A when the distance between the cars is 60 km. on reaching B. it till cover the first quarter of the distance starting from top.. b and c be the distances travelled by the body during xth.. c. each having a speed of 30 km/hr are heading towards each other along a straight path. The acceleration of the particle is (A) c 2b a n2 (B) cba 9 n2 (C) c 2b a 4 n2 (D) c ba n2 3. then the total distance the bird travels till the cars meet is. Choose the correct statement (A) A body starts from rest and moving with constant acceleration travels a distance y1 and the 3rd second y1 5 and y2 in 5th second. A bird that can fly at 60 km/hr flies off car A when the distance between the cars is 60 km. Two cars A and B. heads directly towards car B. second part with constant velocity v and third part with constant retardation a.09 s (B) 0. If a. 3 4 (B) 7 8 (C) 5 6 (D) 9 7 A particle moving in a straight line with uniform acceleration is observed to be at a distance a from a fixed point initially. The ratio y 9 . (A) infinite (B) 30 km (C) 60 km (D) 120 km 6.2 m/s. (A) 25 km/h (B) 54 km/h (C) 40 km/h (D) 50 km/h 16. A target is made of two plates. 2nd. a similar bullet fired with the same velocity from opposite direction goes through iron first and then penetrates 2 cm into wood. A car is moving along a straight road with a uniform acceleration. (A) 2 8 gh g ( t ) 2 (B) g t 8 gh 2 2 (C) 1 2 2 2 8 gh g ( t ) (D) 8 gh 4g2 ( t )2 17. To o Mv increase the velocity of the vehicle from v 1 to v 2. The initial velocity of the ball is. on both days. the time difference between them in reaching the ground in s(g = 10 m/s 2) is (A) 12 (B) 6 (C) 2 (D) 1 15. It was observed at a height h twice with a time interval t..8 m (C) 24. (A) Even numbers (B) Odd numbers (C) All integral numbers (D) Square of integral numbers 14. (A) 4. The thickness of the wooden plate is 4 cm and that of iron plate is 2 cm. Starting from rest a particle moves in a straight line with acceleration a = (25 – t 2)1/2 m/s2 for 0 t 5 s. The velocity of the car midway between P and Q is (A) 33. nth s of its motion are proportional to..9 m (B) 9. (A) Less (B) Same (C) More (D) Cannot be predicted 10. will be. 3 a= m/s2 for t > 5 s. The velocity at any time t is a b (A) v t = b (1 – e–bt) (B) v t = e–bt a 19. 8 (A) 11 m/s (B) 22 m/s (C) 33 m/s (D) 44 m/s 12. A bullet fired goes through the wood first and then penetrates 1 cm into iron.9. The motion of a body falling from rest in a resisting medium is described by the equation a and b are constants.5 m (D) 50 m 13. If the speed of the launch. the time required for the complete journey on the rough day. 3rd. then moves with uniform speed and finally. The velocity of particle at t = 7 s is. If the ratios of the time taken for acceleration. What is the distance between the two bodies 2 sec after the release of the second body. The distance moved by a freely falling body (starting from rest) during 1st. the distance travelled by it (assuming no friction) is A self-propelled vehicle of mass M whose engine delivers constant power P has an acceleration a = (A) s = 3P 2 (v 2 v 12 ) M (B) s = M 2 ( v 2 v 12 ) 3P (C) s = M 3 ( v 2 v 13 ) 3P (D) s = 3P 3 ( v 2 v 13 ) M .. It passes through two points P and Q separated by a distance with velocity 30 km/hr and 40 km/hr respectively. A steam boat goes across a lake and comes back (i) on a quiet day when the water is still. the average speed of the train over the whole journey is. . a (C) v t = b (1 + e–bt) (D) v t = dv = a – bv where dt b bt e a P . A ball is thrown vertically upwards. and (ii) on a rough day when there is a uniform current so as to help the journey onward and to impede the journey back. retards uniformly. as compared to that on the quiet day. A body is released from a great height and falls freely towards the earth. uniform speed and retarded motions are 1 : 8 : 1 and the maximum speed of the train is 60 km/h. one of wood and the other of iron. was same. A ball is thrown vertically upwards with a speed of 10 m/s from the top of a tower 200 m high and another is thrown vertically downwards with the same speed simultaneously. Between two stations a train first accelerates uniformly.3 km/hr (B) 20 3 km/hr (C) 25 2 km/hr (D) 35 km/hr 11. If a1 and a2 be the retardation offered to the bullet by wood and iron plates respectively then (A) a1 = 2a2 (B) a2 = 2a1 (C) a1 = a2 (D) Data insufficient 18. Exactly one sec later another body is released. A river is flowing from west to east at a speed of 5 meters per minute. wants to swim across the river in the shortest time. The velocity is displacement graph of a particle moving along a straight line is shown v v0 x0 The most suitable acceleration-displacement graph will be a a a x (A) x a x (B) (C) x x (D) . A body starts from rest at time t = 0. He should swim in a direction. moving in a semicircle of radius 1. The magnitude of the average velocity (A) 3.) 26. a particle goes from point A to point B. capable of swimming at 10 meters per minute in still water. Neglecting subsequent motion and air resistance. its velocity v varies with the height h above the ground as v v d (A) h (B) v d v d h d h (C) (D) h Acceleration 2 (m/s ) 10 25. The velocity of the river water in km/hr is A (A) 1 (B) 3 (C) 4 (D) 41 23. A boat which has a speed of 5 km/hr in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. A ball is dropped vertically from a height d above the ground.0 m/s (C) 1. A man on the south bank of the river. In 1. (A) 21 m (B) 210 m (C) 2100 m (D) 120 m A 5 B m/s2 20. the acceleration time graph is shown in the figure. (A) due north (B) 30° east or north (C) 30° west of north (D) 60° east of north 22.For an airplane to take-off it accelerates according to the graph shown and takes 12 s to take-off from the rest position. The maximum velocity attained by the body will be (A) 110 m/s (B) 55 m/s (C) 650 m/s (D) 550 m/s Time 11 (Sec. 6 t (in s) 12 21.0 m (see Figure). 0. It hits the ground and bounces up vertically to a height d 2 .0 m B 24. s.14 m/s (B) 2.0 m/s (D) Zero 1. The distance travelled by the airplane is. (B) The two arrows will reach back their starting points at t = 20 s and at t = 25 s. M. Then. Four persons K. The average acceleration in this time is (A) Zero (C) 1 (B) 1 2 m s 2 towards north-east (D) 2 m s 2 towards north-west 1 m s 2 towards north-west 2 28. NO other information is available about its motion at intermediate times (0 < t < 1). If denotes the instantaneous acceleration of the particle. At t = 0. S 4. A particle of mass m moves on the x-axis as follows : it starts from rest at t = 0 from the point x = 0. must be 4 at some point or points in its path. (D) must change sign during the motion. 2. A particle moves in a circle of radius R. L. A projectile is fired with a constant speed at two different angles of projection. Spotlight S rotates in a horizontal plane with constant angular velocity of 0. b(<a) and p are positive constants of appropriate dimensions.5 s. same range. The coordinates of a particle moving in a plane are given by x(t) = a cos(pt) and y(t) = b sin(pt) where a. and x and y are respectively the horizontal and vertical distances of the projectile from the point of projection. Each person now moves with a uniform speed v in such a way that K always moves directly towards L. 2 The trajectory of a projectile in a vertical plane is y = ax – bx .Multiple Choice Question 27. say. The spot of light P moves along the wall at a distance of 3 m. cannot remain positive for all t in the interval 0 t 1. (C) The ratio of the speeds of the first and the second arrows at t = 20 s will be 2 : 1. In half the period of revolution its displacement is _________ and distance covered is ___________. The velocity of the spot P when 45 (see figure) is __________ ms/. A particle is moving eastwards with a velocity of 5 m/s. N are initially at the four corners of a square of side d. 3m P . (D) The maximum height attained by either arrow will be 980 m. The maximum height attained is ________ and the angle of projection from the horizontal is ________. and N directly towards K. an arrow is fired vertically upwards with a speed of 98 ms–1 A second arrow is fired vertically upwards with the same speed at t = 5s. 3. M directly towards N. Then (A) the path of the particle is an ellipse (B) the velocity and acceleration of the particle are normal to each other at t 2 p . 31. and comes to rest at t = 1 at the point x = 1. where a. (C) 29. (A) cosec = sec and and that give it the are such that (B) tan ( + ) (C) sin2 – cos2 = sin2 – cos2 (D) cot = cos sec Fill in the blanks 1. then: (A) (B) cannot exceed 2 at any point in its path. but no other assertion can be made with the information given. The four persons will meet at a time __________. (C) the acceleration of the particle is always directed towards a focus (D) the distance travelled by the particle in time interval t = 0 to t 2 p is a 30. L directly towards M.1 radian/second. In 10s the velocity changes to 5 m/s northwards. b are constants. Then (A) The two arrows will be at the same height above the ground at t = 12. True / False 5. Two balls of different masses are thrown vertically upwards with the same speed. They pass through the point of projection in their downward motion with the same speed (Neglect air resistance). 6. A projectile fired from the ground follows a parabolic path. The speed of the projectile is minimum at the top of its path. 7. Two identical trains are moving on rails along the equator on the earth in opposite directions with the same speed. They will exert the same pressure on the rails. 8. An electric line of forces in the x-y plane is given by the equation x 2 + y2 = 1. A particle with unit positive charge, initially at rest at the point x = 1, y = 0 in the x-y plane, will move along the circular line of force. Table Match 9. Match List I and List II and select the correct answer using the codes given below in the lists: Column-I Column-II I. Deceleration decreasing A. time II. Deceleration increasing B. time velocity III.Acceleration decreasing C. time velocity IV.Uniform acceleration D. time velocity E. time (A) I—D, II—E, III—C, IV—A (C) I—C, II—E, III—B, IV—A (B) I—E, II—B, III—C, IV—D (D) I—D, II—B, III—A, IV—C Passage Type Questions THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE When an airplane flies, its total velocity with respect to the ground is – vtotal v plane vwind Where v plane , denotes the plane’s velocity through motionless air, and vwind denotes the wind’s velocity.. Crucially, all the quantities in this equation are vectors. The magnitude of a velocity vector is often called the “speed”. Consider an airplane whose speed through motionless air is 100 m/s. To reach its destination, the plane must fly east. The “heading” of a plane is the direction in which the nose of the plane points. So, it is the direction in which the engines propel the plane. 1. If the plane has an eastward heading, and a 20 m/s wind blows towards the southwest, then the plane’s speed is – (A) 80 m/s (B) more than 80 m/s but less than 100 m/s (C) 100 m/s (D) more than 100 m/s 2. The pilot maintains an eastward heading while a 20 m/s wind blows northward. The plane’s velocity is deflected from due east by what angle? (A) sin 1 3. 100 m s sin 1 5 (C) tan 1 1 5 (D) None of these (B) 100 m s cos (C) 100 m s (D) sin 100 m s cos Because the 20 m/s northward wind persists, the pilot adjusts the heading so that the plane’s total velocity is eastward. By what angle does the new heading differ from due east? (A) sin 5. (B) cos 1 Let denote the answer to previous question. The plane in question 2 has what speed with respect to the ground? (A) 4. 1 5 1 1 5 (B) cos 1 1 5 (C) tan 1 1 5 (D) None of these Let denote the answer to previous question. What is the total speed, with respect to the ground, of the plane in previous question? (A) 100 m s sin (B) 100 m s cos (C) 100 m s (D) sin 100 m s cos THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE During a car crash, the more rapidly a person decelerates, the more likely she is to be injured. A large deceleration is dangerous, even if it lasts for a short time. Airbags are designed to decrease the magnitude of the deceleration. Before the airbag inflates, the driver continues forward at constant speed. But once the airbag inflates, the driver decelerates gradually, instead of getting thrown into the windshield or steering wheel. 20 v (m/s) Airbag 1 .01 .02 .03 .04 .05 t(s) 20 v (m/s) Airbag 2 .01 .02 .03 .04 .05 t(s) Airbag 3 20 v (m/s) .01 .02 .03 .04 .05 t(s) Three different models of airbags were tested using identical crash test dummies. Sensors measured the velocity of the crash test dummy as a function of time, when the car crashed at 20 m/s into a brick wall. The shown velocity vs, time graphs resulted. Time t = 0 is the moment the car crashes. 6. All three airbags are the same size and shape. Which one inflates most quickly? (A) Airbag 1 (B) Airbag 2 (C) Airbag 3 (D) We cannot determine the answer from the given information. 7. Let amax denote the largest instantaneous acceleration that the crash test dummy experiences during the crash. The best airbag is the one for which (A) Airbag 1 (C) Airbag 3 (B) Airbag 2 (D) We cannot determine the answer from the given information. For airbag 2, which of the following graphs best represents the position of the crash test dummy as a function of time? Let x = 0 m be the dummy’s position at time t = 0s. .01 (A) A 9. time(s) time(s) .01 (B) B D (C) C Position C Position B Position A Position 8. amax is as small as possible. Which airbag is best? .01 time(s) .01 time(s) (D) D For airbag 2, approximately how much distance does the dummy cover between the moment the car crashes and the moment the dummy first makes contact with the airbag? (A) 0.2 m (B) 0.4 m (C) 0.6 m (D) 1.0 m THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Recently, college teams from all over the country sent tennis players to participate in a series of experiments conducted by the Physical Education Department of a major university. A variety of coaching methods was used to improve the players’ serves, described below. Experiment 1 Two groups of 50 tennis players worked on the speed of their basic serves for two weeks. one group consisted solely of right-handed players; the other consisted solely of left-handed players. Half of each group watched videos of a right-handed tennis coach. Each player was told to pattern his or her serve on that of the coach in the video. The players received no verbal or physical guidance. the average speed of each player’s serve was measured at the beginning and end of the two-week period, and changes were recorded in Table 1. Table 1 Players' handedness Right Right Left Left Coach's handedness Right Left Right Left Average change in speed (mph) 5 2 -1 8 Experiment 2 For two weeks, a second group of 100 right-handed tennis players watched the same videos of the righthanded tennis coach. The coach also physically guided 50 of the those players through the motions of the serve. Again, no verbal instruction was given during the experiment. The average speed and accuracy of each player’s serves were recorded at the beginning and end of this two-week period. The results are recorded in Table - 2. (C) People learn more easily by observing someone with similar handedness than by observing someone with handedness opposite their own. (B) Left-handed people are better than right-handed people at imitating the movement of someone with similar handedness. who then physically guided them thought the motions of the serve. Which of the following hypotheses is best supported by the results of Experiment 2– (A) Instructional videos are more helpful for right-handed tennis players than is verbal instruction. Which of the following conclusions could NOT be supported by the results of Experiment– (A) Imitating someone whose handedness is the opposite of one’s own will cause one’s skills to deteriorate. (B) Instructional videos are more helpful for left-handed tennis players than for right-handed tennis players.Table 2 Guided No Yes Average Change in speed (mph) 5 9 Average change in Accuracy 15% 25% Experiment 3 For two weeks. The other 50 players did not observe the video but received verbal instruction from the coach. (D) Physical guidance by a coach improves service accuracy for right-handed tennis players more than for left-handed players. The results of the experiments suggest that the player’s average change in service speed will most closely approximate– (A) –1 mph (B) +5 mph (C) +8 mph (D) +12 mph 14. (D) People learn much better from physical contact plus a visual stimulus than from the visual stimulus alone. Suppose 50 left-handed tennis players watch a video of a left-handed coach and are also physically guided by that coach. (C) Verbal coaching would improve average service speed less than would watching the video. (C) Right-handed coaches are better models for all tennis players than are left-handed coaches. 50 players received no verbal instruction. The results are shown in Table 3. 13. (B) The average service speed of all the players would decrease slightly. they watched the same video of the right handed tennis coach. (C) Physical guidance by a coach improves both speed and accuracy of service for right-handed tennis players. who also physically guided them through the motions of the serve. a third group of 100 right-handed tennis players worked on their basis serves. Table 3 Guided Plus Video Verbal Coaching Average Change in speed (mph) 7 10 10. (D) Right-handed people are better than left-handed people at imitating the movement of someone whose handedness is opposite their own. 12. 11. Which of the following results would the expected if Experiment 3 were repeated using left-handed tennis players and a left-handed coach– (A) The average service accuracy of all the players would increases by at least 30%. . (B) Right-handed tennis players are coached by left-handed coaches more frequently than left-handed players are coached by right-handed coaches. (D) The average service speed of the players who watched the video would increase by at least 8 mph. Which of the following hypotheses is best supported by the results of Experiment 1 alone – (A) Tennis players improve less by observing coaches whose handedness is the opposite of their own than by observing those with similar handedness. 5 . Two balls of identical shape.7 122. Two identical pieces of paper will fall at different speeds. one made of lead and one made of aluminum. assuming it lands at the same height from which it was released. the steel ball falls faster than the feather. Which of the following would Scientist 1 predict? (A) the feather would fall faster than the ball (B) the ball and feather would fall at the same speed. a crumpled piece of paper falls at a different speed from a flat piece of paper because of its– (A) mass (B) air resistance (C) gravitational attraction (D) texture 18. Scientist 2 : The mass of an object does not determine how fast an object falls. According to Scientist 1. 21. Since the shape of the steel ball gives it less air resistance than a feather. but shape does. According to Scientist 2. Air buoys up objects falling through it. Table 1 Trial 1 2 3 4 Launch speed (m/s) 10 20 30 40 Range (m) 8 31. Suppose a small ball and a feather having the same weight are dropped from the same height. (C) Scientist 2: The lead ball falls faster because the balls have the same air resistance. steel ball 17. if one of them is crumpled into a small ball.THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE It is known that if a steel ball and a feather are dropped from the same height.8 70. The steel ball falls faster because it is more massive than the feather. The “range” is the distance between where the ball lands and where it was released. Scientist 2 would predict that in a vacuum. and thus is attracted more strongly by the earth’s gravitational field. (C) the ball would fall faster than the feather (D) none of the above. Which scientist’s prediction and reasoning is incorrect based on the arguments presented? (A) Scientist 1 : the lead ball falls faster because the balls have different masses. and launches it at the same angle. In each trial. The greater the gravitational force on an object. Table 1 shows the experimental results. 15. According to Scientist 1. are both dropped from the same height. the property that determines how fast an object will fall is its– (A) chemical composition (B) mass (C) shape (D) gravity 16. it falls faster. The following scientists have two different views on falling bodies: Scientist 1 : The force of gravity makes things fall. two objects would fall at the same speed if they had the same– (A) shape and different masses (B) air resistance and shape (C) composition and air resistance (D) mass and different shapes 20. That is because gravity is not the only force acting on a falling object. (D) Scientist 2: The balls fall at the same rate because they have the same air resistance. the student uses the same baseball. the speed of the falling ball could be increased by– (A) dropping the ball from a greater height (B) making the ball out of aluminum (C) reshaping the ball (D) using a larger. Both scientists agree that the rate at which an object falls is affected by the– (A) force of gravity (B) mass of the object (C) object’s resistance of air (D) shape of object 19. the faster it falls. THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE A student performs an experiment to determine how the range of a ball depends on the velocity with which it is released. (B) Scientist 1 : the lead ball falls faster because the balls have different gravitational attraction. 27. which of the following graphs best represents the vertical component of the ball’s velocity as a function of time. depends on the initial speed. as measured from the horizontal. if the stone is raised above the stake by only the thickness of a leaf. the student then hypothesizes that the range. Thus. assuming upward is positive? + vy (A) 0 + vy time (B) 0 + vy time (C) 0 + vy time (D) 0 time THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Aristotle developed a system of physics based on what he thought occurred in nature. Aristotle also noted that stones dropped into water continue to fall. the stake will be driven in a much smaller amount. R. the ball’s speed is– (B) 0 sin (A) 0 (C) 0 cos (D) 0 26. A 5–pound stone. Galielo disagreed with Aristotle’s explanation. The student speculates that he constant C depends on: I.0 meters 25. For trial 2. the stake will be driven into the ground.0 meters (D) 4. Its range is approximately– (A) 1.0 meters (B) 2. all object exactly the same acceleration. To account for this. say. Consider a stake partially driven into the ground and a heavy stone falling from various heights onto the stake. falls with a constant speed 5 times as great as that of a 1-pound stone. the distance the object falls varies directly with the square of the time of fall if the effect of air resistance on the object is negligible. according to the following to the following equation: R Cv0n . he explained that the resistance of the medium through which an object falls also affects the speed.Based on this data. For example. where C is a constant. II. 4 finger breadths. The angle at which the ball was launch. Let denote the angle of the ball’s initial velocity. Based on this data. Aristotle. and this resistance is the same for all objects. II and III. 22.0 m/s. 0 . 24. Galileo argued. Also. Which graph accurately represents Galileo’s theory of the relationship between speed and time for an object falling from rest under conditions of negligible air resistance – .0 meters (C) 3. The ball’s mass III. If the stone falls from a height of 4 cubits. for example. On the basis of a careful set of experiments. Neglect air resistance. according to Galileo. he said. Therefore. Galielo argued that the speed of an object released from rest varies directly with the time of fall. it instantaneously reaches a speed that remains constant as the stone falls. and n is another constant. He also believed that the speed attained by a stone falling in air varies directly with the weight of the stone. and if air resistance is negligible. objects actually fall with constant acceleration. But if the stone falls from a height of 1 cubit. then the effect of the stone’s falling on the stake will be altogether unnoticeable. he thought that if a stone is released from rest. the best guess for the value of n is– (A) 1 2 (B) 1 (C) 2 (D) 3 23. but at a slower rate than stones falling through air. He generated the following arguments to refute. At the peak (highest point) of its trajectory. The ball’s diameter If we neglect air resistance. The student performs another trial in which the ball is launched at speed 5. the speed of a falling object also varies inversely with the resistance of the medium. Certainly. then C actually depends on– (A) I only (B) I and II (C) I and III (D) I. just when the train starts to move. if air resistance were negligible. (B) The separation between the 2 pieces is constant until the strike the ground. how many seconds would it take an object half as heavy to fall to the ground from the same height– (A) 0. Path of ball 1 seen by Sweta will be (A) Parabolic (B) Straight line (C) Circle (D) Can’t predict 32.5 t (D) 2. To be consistent with Aristotle’s views. 1 second later. According to Aristotle. and the 2 piece travel together with a speed faster than the speed of either. it gets deviated by 20 cm. from what height. Suppose a heavy object falls to the ground in t seconds when dropped from shoulder height. The path of ball 2 as seen by Sweta will be (A) Parabolic (B) Straight line (C) Circle (D) Can’t predict 35. 50 m 50 m Initial Final Now Abhishek drops ball 2 at the same time (t = 50 sec. and the 2 pieces travel together with the speed of the heavier piece. Path of ball 1 seen by Abhishek will be – (A) Parabolic (B) Straight line (C) Circle (D) Can’t predict 33. A piece of putty weighing 2 pounds is dropped down a shaft from the top of a tall building. what happens to the 2 pieces of putty if they fall for a relatively long time– (A) The separation between the 2 piece constantly increases until they strike the ground.5 t (B) 1. The speed with which Abhishek throw ball 3 will be (A) 30 (B) 20 (D) Can’t predict (C) 15 (D) None of these . For 50 seconds train accelerates and attains the maximum speed and starts moving with this velocity. A book dropped from a height of 1 meter falls to the floor in t seconds. THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Abhishek and Sweta Bachhan are two friends Abhishek is Joint to IIT-Pawai and Sweta come to see off him at Indore railway station. The path of ball 2 as seen by Abhishek will be (A) Parabolic (B) Straight line (C) Circle (D) Can’t predict 36. Non Abhishek drops ball 1 from 1 mt height. According to Galileo.) Abhishek throw ball 3 such that it just reach Sweta. (D) The heavier piece catches up to the smaller piece. The acceleration of train is (A) g/2 (B) g/4 (C) g/5 34. in meters. should a book 3 times as heavy be dropped so that it will fall to the floor in the same amount of time– (A) 1 9 (B) 1 3 (C) 1 (D) 3 29. Abhishek observes that ball doesn’t hit the floor point exactly below the dropped point.0 t 30.0 t (C) 1.Time Time (D) Speed (C) Speed (B) Speed Speed (A) Time Time 28. 31. a 3 pound piece of putty si dropped down the shaft. (C) The heavier piece catches up to the smaller piece. his initial kinetic energy. At this angle. what is the expression for minimum length of the cable so that he doesn’t hit is head on the ceiling as he swings upward? 2v02 (A) g v02 (B) g v02 (C) 2g v02 (D) 4g . From Figure 2.4 s (C) 5. approximately how much time will it take for clown with a mass of 60 kg to reach the safety net located 10 m below the height of the cannon? (A) 4. (B) moved farther away from the cannon and raised. The cannon will accelerate the clown by applying a constant force of 10. A large hoop is to be suspended from the ceiling by a massless cable at just the right place so that the clown will be able to dive through it when he reaches a maximum height above the ground. 42 vy (m/s) 28 21 80 45° 60 kg 40 kg kg t(s) 2. mv02 2 . what initial velocity v0 will he have as he leaves the cannon? (A) 3m s (B) 15 m s (C) 30 m s (D) 300 m s 39. the hoop will have to be: (A) moved farther away from the cannon and lowered. After passing through the hoop he will then continue on his trajectory until arriving at the safety net. If the angle the cannon makes with the horizontal is increased from 45°. A clown will be shot out of a cannon with velocity v0 at a trajectory that makes an angle 45 with the ground. (D) moved closer to the cannon and raised. If the mass of a clown doubles. If the clown’s mass is 80 kg.2 Figure 2 Figure 2 shows a graph of the vertical component of the clown’s velocity as a function of time between the cannon and the hoop.THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE A circus wishes to develop a new clown act. The slope of the line segments plotted in figure 2 is a constant. Since the velocity depends on the mass of the particular clown performing the act.9 s (D) 7. Figure 1 38. Figure 1 shows a diagram of the proposed setup. will: (A) remain the same (C) double (B) be reduced to half (D) four times 42. 37.1 2.8 4. The height above he ground at which the clown begins his trajectory is 10 m. (C) moved closer to the cannon and lowered. If a clown holds on to the hoop instead of passing through it. the graph shows data for several different masses.2 s 41. the clown will travel a maximum horizontal distance.000 N over a very short time of 0. Which one of the following physical quantities does this slope represent? (A) g (B) v0 (C) y y0 (D) sin 40.24 sec.3 s (B) 6. If the velocity of the particle is 0. m/s 3 0 0 2.Level # 2 M O TI O N I N O N E D I M E N SI O N 1. prove that the total time taken is t = shortest time ? v 1 1 . Determine the velocity and displacement as functions of time. At what value of v is the time of travel the shortest ? What is the value of the v 2 6. including d = 30 m and the critical separation distance dc.. choosing x = 0 at t = 0. A particle starts from rest at x = –2 m and moves along the x-axis with the velocity history shown. A base ball player hits a pitched ball so that its velocity reverses direction and its speed changes from 30 m/s to 40 m/s.2 0 9. 0.m/s x 8. Calculate the speed of the point of intersection of the lines if the angle between them is . The speed of a train increases at a constant rate from zero to v and then remains constant for an interval and finally decreases to zero at a constant rate . The acceleration of a particle that moves in the positive x-direction varies with its position as shown. The launch travelled 4. and it is in contact with the ball for a distance of 0. Find the time required for the car to travel 400 m from the start. If is the total distance described. The bat moves at an average velocity of 30 m/s.4 0. (A) If d = 30 m and he keeps running. 0 0.2 v. When the launch began traveling upstream.5 1.8 1. The speed of a motor launch with respect to the water is v = 7 m/s. a .4 0. How long is it before the launch reaches the float again ? Assume that float is moving with the speed of stream. after which car continues at a constant speed. 5. A passenger is running at his maximum velocity of 8 m/s to catch a train. On the same graph sketch x (t) for the passenger for various value of initial separation distance d. Plot the corresponding acceleration and displacement histories for the two seconds.5 -1 . Find the time t when the particle crosses the origin. (C) For the critical separation distance d0. A car starts from rest with an acceleration of 6 m/s2 which decreases to zero linearly with time.4 m. The acceleration of a particle is given by a = 4t – 30. determine the velocity v of the particle when x = 1. and the initial velocity is v 0 = 3 m/s. 4. a float was dropped from it. 2. turned about and caught up with the float. in 10 sec.0 1. where a is in meters per second squared and t is in seconds.0 0. will he be able to jump onto the train. the speed of the stream u = 3 m/s. (A) For how long are the bat and ball in contact ? (B) What is the average acceleration of the ball while it is in contact with the bat ? 3.05 m. When he is a distance d from the nearest entry to the train. Two intersecting straight lines moves out parallel to themselves with the speed v 1 and v 2.2 km upstream. such that he just catches the train. (B) Sketch the position function x(t) for the train. the train starts from rest with constant acceleration a = 1 m/s2 away from the passengers. what is the speed of the train when the passenger catches it ? What is its average speed for the time interval from t = 0 until he catches it ? What is the value of dc.8 m/s when x = 0. The initial displacement at t = 0 is 2 s0 = –5m. 7. (A) What is the velocity of the ball relative to a stationary observer by the side of the read ? (B) What is the velocity of the ball relative to the driver of a car moving in the same direction as the truck at a speed of 90 km/h. At a certain instant. Another student standing. A man running on a horizontal road at 8 km/h. The velocity of a ship in still water is 20 km/h. Also find the distance travelled by the helicopter during the interval of dropping the packets. Determine (A) When and where the ball will hit the elevator. Find the speed and the direction of the rain with respect to the read. when second packet is dropped. At the same instant an open platform elevator passes the 5m level. ? . that the motor boat heads towards the ship at 60° ? 19.25 s after it fell off ? 16. another food packet is dropped from it. level road throws a ball with a speed of 20 km/h relative to the truck in the direction opposite to its motion. An object is thrown upward with an initial velocity v 0. 11. As this packet strikes the ground. Two bodies moves in a straight line towards each other at initial velocities v 1 and v 2 and with constant accelerations a1 and a2 directed against the corresponding velocities at the initial instant. A ball is thrown vertically upward from the 12 m level in an elevator shaft with an initial velocity of 18 m/s. 14. moving upward with a constant velocity of 2 m/s. (B) the relative velocity of the ball with respect to the elevator when the ball hits the elevator. between the bodies for which they meet during the motion ? 18. On a straight. The drag on the object is assumed to be proportional to the velocity. 20. (B) How much later does the heavier iron ball strike the ground ? 12. At what fraction of the height of the building did the collision occur ? 13. RELATIVE MOTION 17. When balls collide. which takes 4 seconds to reach the ground. find the rain falling vertically. (A) Neglecting air resistance. What time will it take the object to move upward and what maximal altitude will it reach ? 15. What is the velocity of a motor boat approaching the ship at right angle to its course if it appears to people on board the ship. A helicopter is descending vertically downward with a uniform velocity.MOTION UNDER GRAVITY 10.8 m/s. He increase the speed to 13 km/ h and find the drops make angle 30° with the vertical. One student throws a ball vertically upward with an initial speed of 9. 5m away starts running towards the ball on release and catches it at the same height. a food packet is dropped from it which takes 5 seconds to reach the ground. A nut comes loose from a bolt on the bottom of an elevator as the elevator is moving up the shaft at 3 m/s. An aluminium ball with a mass of 4 kg and an iron ball of the same size with a mass of 11. (B) How far above the bottom was the nut 0. (A) How far from the bottom of the shaft was the elevator when the nut fell off. A person riding in the back of a pickup truck traveling at 60 km/h. Find the velocity with which the helicopter is descending and its height. they are moving in opposite directions and the speed of A is twice the speed of B. Ball A is dropped from the top of a building at the same instant that ball B is thrown vertically upward from the ground. What must be the maximum initial separation max.6 kg are dropped simultaneously from a height of 49 m. What was the students acceleration ? (Assume a uniform acceleration). The nut strikes the bottom of the shaft in 2 sec. how long does it take the aluminium ball to gall to the ground. After the collision P retraces its 135° path. Their inclinations to the horizontal being and . A body is dropped from a stationary balloon at a height h above the ground. 24. OC. OC. The initial velocities of P and Q makes 45° and 135° angles respectively with the horizontal as shown in the figure. OB. If the truck maintains a constant speed u = 15 m/s.8 m/s2. Both particles travel in the same vertical plane and under go a collision. determine the position s where the ball strikes the top of the truck. If the angle of elevation of the balloon from the position of the gun is . Determine the position of Q when it hits the 45° ground. Find the time taken by each boy to reach at the point A. A boy throws a ball horizontally with a speed of v 0 = 12 m/s from the gandhi setu bridge c of patna in an effort to hit the top surface AB of a truck travelling directly underneath the boy on the bridge. v0=12 m/s 8m s u B A C 10m u 26. D and E. Derive a formula for a time needed to make a round trip of total distance D if the boat makes the round trip by moving (A) Upstream and back down stream. in which direction should the bullet be fired so that it strikes the body before reaching the ground. Particles P and Q of mass 20 gms and 40 gms respectively are simultaneously projected from points A and B on the ground. OD and OE simultaneously. How much time after the collision does the A B 245 m particle Q take to reach the ground. Also find the minimum value of u required for this.21. OB. C. Why ? O 22. The separation AB is 245 m. P a O . (B) Directly across the river and back. 25. then find the velocity of projection. We must assume u < v . OD and DE is shown. If a particle is projected at right angle to the former from a point in it so as to strike the other at right angles. OGC is the diameter of the circle. Each particle has in initial speed of 49 m/s. B. Two inclined planes intersect in a horizontal plane. G A E D C B PROJECTILE MOTION 23. and the ball is projected at the instant B on the top of the truck appears at point C. a vertical cemented circular plane of radius R having frictionless slots along the chords OA. At the same time a bullet is fired from a gun on the ground with a velocity u. Five Boys starts sliding down from rest along the slots OA. The boat is to make a round trip in a river whose current travels at speed u. The speed of a boat in still water is v. Take g = 9. In the adjacent figure. at an elevation of 30°. 3 (A) Calculate the trajectory of the particle. A juggler manages to keep five balls in motion. 34.) 29. (B) Find its distance from the origin at time . i. where k is a positive constant. 28.27.6 ˆi 6. A bomb bursts on contact with the ground and pieces fly off in all directions with speed up to 30 m/s. A boy sitting at the rear end of a railway compartment of a train running at a constant acceleration a on horizontal rails fires a shot towards the fore end of the compartment with a muzzle velocity u = 20 m/s at an angle = 37° above the horizontal when the train’s velocity v = 10 m/s. (B) Sketch the path of the projectile as observed by an observer outside the elevator. A ball is shot from the ground into the air. Also.1ˆj in water per second ( ˆi horizontal. F = – kv. u u0 ˆi a cos t ˆj . (a) Determine the time interval between successive throws. What is the time duration over which she can be hit by a piece ? 31. If the boy catches the shot without moving from his seat at the same height as that of projection find (a) Speed of the train at the time when he catches the shot and (B) the acceleration of the train (g = 10 m/s2). A cannon fires from under a shelter inclined at an angle to the horizontal.e. a stone is projected upward from its floor with a speed of 2 m/s relative to the elevator. v0 0 x An elevator is going up with an upward acceleration of 1 m/s2. The initial velocity of the shell is v 0 and its trajectory lies in the plane of the figure. . where ˆi and ˆj are unit vectors along x and y axes respectively. At a height of 9. throwing each sequentially up a distance of 3m. its velocity is observed to be v 7.1 m. 200 A man is travelling on a flat car which is moving up a plane inclined at cos–1 (4/5) to the horizontal with a speed 5 m/s. how would the motion be altered ? A particle is moving in the plane with velocity given by. If the liquid develops a frictional or drag resistance on the projectile which is proportional to its velocity. A girl is standing 40 m away. (A) Calculate the time taken by the stone to return the floor. At the instant when its velocity is 2 m/s. (Neglect the time taken to transfer balls from one hand to the other. that it travels ? 33. ˆj upward). A projectile of mass m is fired into a liquid at an angle 0 with an initial velocity v 0 as shown. If the distance d of the hoop from the level of the man’s hand is such that its component perpendicular to the incline is 4 m.. (A) (B) (C) (D) To what maximum height does the ball rise ? What total horizontal distance does the ball travel ? What are The magnitude and The direction of the ball’s velocity just before it hits the ground ? 30. calculate the time taken by the ball to reach the hoop. He throws a ball towards a stationary hoop located perpendicular to the inclined in such a way that the ball moves parallel to the slope of the incline while going through the hoop. 35. determine the x and y components of its velocity at any instant. If particle is at the origin at t = 0 . (B) Give the positions of the other balls at the instant when one reaches her hand. what is the maximum distance x max. y 32. (C) If the elevator was moving with a downward acceleration equal to g. The canon is located at a point at a distance from the base of the shelter.. Determine the maximum range of the shell. assumed to be the x-y plane. 42. Find the maximum distance of the projectile from the inclined plane. find when and where the bullets may meet. At a particular instant. On a frictions horizontal surface. If the guns are shot with velocities of 350 m/s upward and 300 m/s downward respectively. Two guns are pointed at each other one upwards at an angle of elevation 30° and the other downwards at the same angle of depression. one upward at an angle of 30 ° and the other at the same angle of depression. Its velocity makes an angle with the x-axis and it hits the trolley. A 30 m P y 30° x Bomber 40. (a) The motion of the ball is observed from the frame of the trolley. Calculate the angle made by the velocity vector of the ball 45° O x with the x-axis in this frame. If the charges leave the guns with velocities 400 m/s and 300 m/s respectively. ? 37. For what value of the angle will the water land closest to the wall after clearing the top ? Neglect the effects of wall thickness and air resistance. Find the angle and the speed of projection of the bullet. Where does the water land ? 38. a ball is thrown along the surface from the origin O. V0 P1 P2 O 30 m 20 m 20 m 170 m B 30° 39. if 4 . a small trolley A is moving along a straight line parallel to the y-axis (see figure) with a constant velocity of y 3 1 m s .36. Water is ejected from the water nozzle with a speed v 0 = 14 m/s. Determine the angle of the line of sight of the bomber with the ship at the instant a bomb is released so as to hit the ship. 3 [IIT 2002] . Where would the bomber be at the instant the ship is wrecked ? Line of Flight Trajectory 3 km Line of Sight ship 41. (B) Find the speed of the ball with respect to the surface. 43. If the line connecting the man and the object makes an angle of 30° to the train at the instant of shooting to what angle to the train should be aim in order to hit the object if the muzzle velocity is 850 km/hr. find when and where they meet. the muzzles being 42 m apart. A bomber is flying horizontally at a speed of 500 km/h at an altitude of 3 km such that a ship lies in a vertical plane through the line of sight as shown in figure. the muzzles being 30 m apart as shown in figure problem. Two guns are projected at each other. A rifleman on a train moving with a speed of 60 km/hr. A particle is projected up an inclined plane of inclination with an initial velocity u at an angle to the horizontal. A bullet is projected so as to graze the tops of two walls each of height 20 m located at distances of 30 and 170 m in the same line from the point of projection as shown in figure. fires at an object running away from the train at right angles with a speed of 45 km/hr. when the A line OA makes an angle of 45° with the x-axis. Q. [IIT 1996] Take origin of the coordinate system at the foot of the hill right below the muzzle and trajectories in x-y plane. A n s. 0. One gun fires horizontally and other fires upwards at an angle of 60° with the horizontal. 1 A 10 C 19 C 28 AD Que. d v 6. The shots collide in air at a point P.6 7. A 9 C 18 A 27 B . An object of mass m is thrown from the top of AB horizontally with a velocity of 2m 10 m s towards CD. The two objects move in the same vertical plane. Simultaneously another object of mass 2 m is thrown from the top of CD at an angle of 60° to the horizontal towards AB with the same magnitude of initial velocity as that of the first object. A d B D Answer Key Assertion & Reason Q. 45. (ii) Find the position where the objects hit the ground. C D A A A A C B D E E 2 A 11 B 20 B 29 ABC 3 A 12 C 21 A 30 ABC L ev e l – 1 4 5 A C 13 14 B C 22 23 B B 31 ABD 6 D 15 B 24 A 7 B 16 C 25 B 8 AB 17 B 26 A Fill in the Blanks / True–False / Match Table 1. A n s. Find (a) the time-interval between the firings. situated on the top of a hill of height 10 m. 2 r . fire one shot each with the same speed 5 3 ms–1 at some interval of time. collide in mid-air and stick to each other. F C 60° 4. Q. Two towers AB and CD are situated a distance d apart as shown in figure AB is 20 m high and CD is 30 m high from the ground. A n s. a 2 4b . T 3. (i) Calculate the distance ‘d’ between the towers and. Two guns.44. 8. 5. F tan 1 a 9. and (b) the coordinates of the point P. A n s. Q. T r 2. 1 2 3 4 5 6 7 8 9 10 11 Ans. g sin 2 sin v0 .5 m. v min. t = 0.04 m/s downward at 177.2 m downward. (a) DV (b) v u2 2 19. Ans. = gh . t m = r n 1 g v2 . 122.6 m (b) 14 m 16.917 s MOTION UNDER GRAVITY 10. 25. Ans.166 m/s 2. every boy will take same time PROJECTILE MOTION 23. Que.cosec 2 2 ag 26. = ( v u) 15 t 2 + 2/3 t 3 8.65 s. Ans. (a) 0. 2/3 13. 3. 55. h = 0 2g 12.001667 s (b) 42000 m/s2 3.30 m (b) – 19.8 k height. cos ec 4. ( ) 2 2 5. 16. 3. (a) 13. u sin sin sin cos( ) 24. t = 2 18. V = 1. 4 m/s 32 m 7. v = 3 – 30t + 2t 2. .84 m v 02 1 g sin 2 27.6 km/h 21.KINEMATICS Passage Type Que. 11. (a) 3.2 s. O 1 r v0 14. Que.81 m/s 15. 12.5 m/s2 11. max ( v 1 v 2 )2 2 (a1 a 2 ) 20. 3. Ans. 2.678 6. (a) yes (b) 8 m/s. 4 7 km/h D 2 v u2 R g .53 sec. RELATIVE MOTION 17. 1 B 12 C 23 A 34 A 2 C 13 D 24 B 35 B 3 D 14 A 25 C 36 D 4 A 15 B 26 A 37 D 5 A 16 D 27 A 38 C 6 A 17 B 28 D 39 A 7 B 18 A 29 B 40 C 8 C 19 B 30 D 41 B 9 A 20 B 31 B 42 D 10 D 21 C 32 B 11 A 22 C 33 C Level # 2 M O TI O N I N O N E D I M E N SI O N 1. 34. V v 12 v 22 2 v 1v 0 cos . t min. S = –5 + 3t – 9. Que. (a) 40 km/h (b) – 50 km/h 22. 5' 38.547 m from B —X—X—X—X— (b) 5 3 m.6 sec m k k / m t g .068 s. 11. In 0. (a) 310 m/s (b) 1. s u 2 sin 2 ( ) 2g cos 43.78 and 11.9 m above the hands 29. (a) 11m (b) 23 m (c) 17 m/s (d) 63 below horizontal. v x = v 0 cos . 34°. 42 m/s 33. 4.07 m) 40.28. 44.18 sec 35. over the ship.0462 s. 8. 41. = 50. at a point whose horizontal of vertical dist. 30. v y = k m v 0 sin g e k m x 34.32. 48.7°. 5m . 36. (a) y = a sin u (b) 0 9 2 u02 a 2 4 2 37. 17.98 m 42. 0.9°.9 m and 2.835 to the right of B 31. e(– km) t. (14 m. 48 m/s 39. from the 1st gun are 20. 0. (a) 45° (b) 2 m/s. (a) t 1 sec ond 45. (a) 0. x = mv 0 cos 0 32. 1 sec.