Physics P Worksheet 9.1 Momentum and Impulse Worksheet 9.1 Momentum and Impulse 1. The momentum of a 3000 kg truck is 6.36 × 104 kg·m/s. At what speed is the truck traveling? 2. A 0.40 kg object is moving on a frictionless surface with a speed of 30 m/s. A force of 2.0 N is applied continually until the velocity of the object has been reversed. How long was the force applied? 3. A 0.058–kg golf ball is struck by a club that produces a force of 272 N and gives the ball a velocity of 62 m/s. For how long was the club in contact with the ball? 4. A force of 186 N acts on a 7.3–kg bowling ball for 0.40 s. What is the change in velocity of the bowling ball? 5. Small rockets are used to make tiny corrections to the speed of a satellite. If a rocket has a thrust of 35 N and is used to change the velocity of a 72000–kg satellite by 63 cm/s, for how long should it be fired? 6. A tennis player practices against a backboard, hitting the ball to the board with a speed of 22 m/s. The ball bounces straight back from the board with a speed of 19 m/s. The mass of the ball is 55 g. What is the average force the wall exerts on the ball if the ball is in contact with the board for 12.5 ms (0.0125 s)? 7. A force of 186 N acts on a 7.3–kg bowling ball for 0.40 s. a) What is the bowling ball’s change in momentum? b) What is its change in velocity? 8. A 0.24–kg volleyball approaches Tina with a velocity of 3.3 m/s. Tina bumps the ball, giving it a speed of 2.4 m/s but in the opposite direction. What average force did she apply if the interaction time between her hands and the ball was 0.025 s? 9. Before a collision, a 25–kg object was moving at +11 m/s. Find the impulse that acted on the object if, after the collision, it moved at the following velocities: a) +8 m/s b) –3 m/s 10. A nitrogen molecule with a mass of 4.7 × 10-26 kg, moving at 550 m/s, strikes the wall of a container and bounces back at the same speed. a) What is the impulse the molecule delivers to the wall? b) If there are 1.5 × 1023 collisions each second, what is the average force on the wall? 0$N Δt = 12$s 3.36$ × $10 4 $kg⋅m/s2 v= 3000$kg v = 21. Δp F= Δt Δp Δt = F p f − pi Δt = F Δt = ( m v f − vi ) F Δt = ( )( 0.40$kg −30$m/s − 30$m/s ) +$2. p = mv p v= m 6.1 Momentum and Impulse 1.013%s 4.2$m/s 2.3$kg Δv = 10$m/s .Physics P Worksheet 9.40$s ) 7. Δp F= Δt Δp Δt = F p −p Δt = f i F Δt = ( m v f − vi ) F Δt = ( )( 0. Δp F= Δt Δp = FΔt mΔv = FΔt FΔt Δv = m Δv = ( )( 186$N 0.058%kg 62%m/s − 0%m/s ) 272%N Δt = 0. 63$m/s)( ) 35$N Δt = 1300$s 6.3#kg Δv = 10#m/s 8. Δp F= Δt F= ( m v f − vi ) Δt F= ( )( 0.1 Momentum and Impulse 5. Δp F= Δt F= ( m v f − vi ) Δt F= ( )( 0.40$s ) Δp = 74$kg⋅m/s 7b.Physics P Worksheet 9. Δp = mΔv Δp Δv = m 74#kg⋅m/s Δv = 7. Δp F= Δt Δp = FΔt ( )( Δp = 186$N 0.00125$s F = '1800$N 7a.24%kg (2. Δp F= Δt Δp Δt = F mΔv Δt = F Δt = ( 72000$kg 0.055$kg '19$m/s$'$22$m/s ) 0.3%m/s ) 0.4%m/s%(%3.025%s F = (55%N . 5 × 1023 collisions per second then each collision lasts an average of 1/1. If there are 1. Δp = mΔv ( Δp = m v f − v i ) ( )( Δp = 4.5 × 1023 s.2$ × $10 $kg⋅m/s 10b.7$ × $10 −24 $s F = 7.2$ × $10'23 $kg⋅m/s F= 6.1 Momentum and Impulse 9a. Δp F= Δt 5.Physics P Worksheet 9. Δp = mΔv ( Δp = m v f − v i ) ( )( Δp = 25#kg 8#m/s#*#11#m/s ) Δp = *75#kg⋅m/s 9b.8$N . Δp = mΔv ( Δp = m v f − v i ) ( )( Δp = 25#kg &3#m/s##m/s ) Δp = &350#kg⋅m/s 10a.7$ × $10 −26 $kg +550$m/s$+$550$m/s ) −23 Δp = +5.