Impulse Momentum Problems Answers 1. A force of 6.00 N acts on a 3.00-kg object for 10.0 s. (a) What is the object’s change in momentum? F t = ∆p F = 6.00 N t = 10.0 s = ( 6.00 N )( 10.0 s ) = ∆p = 60.0 N s (b) What is its change in velocity? p = m v ∆p = m ∆v m m ∆p 60.0 N s ∆v = = Challenge: Show that m 3.00 kg N s = m/s kg ∆v = 20.0 m/s

2. What force is needed to bring a 1500-kg car moving at 22.0 m/s to a halt in 20.0 s? p = m v = ( 1500 kg )( 22.0 m/s) = 33 000 kg m/s pi = 33 000 kg m/s pf = 0 ∆p = pf - pi F t = ∆p = 0 - 33 000 kg m/s t t ∆p = - 33 000 kg m/s ∆p F = t - 33 000 kg m/s = 20.0 s F = - 1650 N

3. Calculate the momentum of a baseball of mass 0 3. Calculate the momentum of a baseball of mass 0.15 kg when it is traveling at a speed of 46 m/s. p = m v = ( 0.15 kg )( 46 m/s ) p = 6.9 kg m/s

4. A car of mass 1000 kg is accelerated from rest at a rate of 3 4. A car of mass 1000 kg is accelerated from rest at a rate of 3.0 m/s2 for a time of 6.0 s. Calculate its momentum at the end of this time. p = m v v = ? m = 1000 kg a = 3.0 m/s2 t = 6.0 s vi = 0 vf = vi + a t = 0 + ( 3.0 m/s2 )( 6.0 s ) vf = 18 m/s p = m v = ( 1000 kg )( 18 m/s ) p = 18 000 kg m/s

5. A car weighing 15 680 N and moving at 20.0 m/s is acted upon by a 640 N force until it is brought to a halt. (a) What is the car’s mass? Wt. = m g g g Wt. 15 680 N = m = 1600 kg m = = g 9.8 m/s2

5. A car weighing 15 680 N and moving at 20.0 m/s is acted upon by a 640 N force until it is brought to a halt. (b) What is its initial momentum? pi = m vi = ( 1600 kg )( 20.0 m/s ) pi = 32 000 kg m/s (c) What is the change in the car’s momentum? ∆p = pf - pi pf = 0 = 0 - 32 000 kg m/s ∆p = - 32 000 kg m/s

5. A car weighing 15 680 N and moving at 20.0 m/s is acted upon by a 640 N force until it is brought to a halt. (d) How long does the braking force act on the car to bring it to a halt? F t = ∆p F = - 640 N (braking force) F F ∆p - 32 000 kg m/s - 32 000 kg m/s t = = = F - 640 N - 640 kg m/s2 t = 50 s

6. A tennis ball of mass 55.0 g is traveling towards Bob at 25.0 m/s. After hitting it, it travels away from Bob at 30.0 m/s. (a) What is the change in velocity? - 25.0 m/s + 30.0 m/s vi = - 25.0 m/s vf = + 30.0 m/s ∆v = vf - vi = ( + 30.0 m/s ) - ( - 25.0 m/s ) ∆v = + 55.0 m/s

6. A tennis ball of mass 55.0 g is traveling towards Bob at 25.0 m/s. After hitting it, it travels away from Bob at 30.0 m/s. (b) If the ball was in contact with the racket for 0.15 s, what force did the racket exert on the ball? ∆v = + 55.0 m/s F t = ∆p = m ∆v m = 55.0 g = 0.055 kg t t m ∆v ( 0.055 kg )( 55.0 m/s) F = = t 0.15 s F = 20.2 N