Physics Chapter 9 | Collision | Mass

# Chapter 9 Impulse and Momentum

Physics, 6th Edition

Chapter 9. Impulse and Momentum
9-1. A 0.5-kg wrench is dropped from a height of 10 m. What is its momentum just before it strikes the floor? (First find the velocity from conservation of energy.) mgh = ½mv2; v = 2 gh = 2(9.8 m/s 2 )(10 m) v = 14.0 m/s

p = mv = (0.5 kg)(14 m/s);

p = 7.00 kg m/s, down

9-2. Compute the momentum and kinetic energy of a 2400-lb car moving north at 55 mi/h. m= W 2400 lb = ; m = 75 slugs ; v = 55 mi/h = 80.7 ft/s g 32 ft/s 2 p = 6050 slug ft/s K = 244,000 ft lb

p = mv = (75 slugs)(80.7 ft/s); K = ½mv2 = ½(75 slugs)(80.66 ft/s)2;

9-3. A 2500-kg truck traveling at 40 km/h strikes a brick wall and comes to a stop in 0.2 s. (a) What is the change in momentum? (b) What is the impulse? (c) What is the average force on the wall during the crash? Take + to be toward the wall. ( 40 km/h = 11.1 m/s) ∆ p = mvf – mvo = 0 - (2500 kg)(11.1 m/s); ∆ p = - 27,800 kg m/s

Impulse = ∆ p; F t = -27,800 kg m/s Force ON truck: F= −27,800 ; 0.2 s F = -139,000 N F = + 139,000 N

Force on wall is opposite, so

9-4. What is the momentum of a 3-kg bullet moving at 600 m/s in a direction 300 above the 600 m/s horizontal? What are the horizontal and vertical components of this momentum?
300

p = mv = (3 kg)(600 m/s); px = 1800 cos 300 and py = 1800 sin 300;

p = 1800 kg m/s, 30

0

px = 1560 kg m/s; py = 900 kg m/s

101

Chapter 9 Impulse and Momentum

Physics, 6th Edition

*9-5. A 0.2-kg baseball traveling to the left at 20 m/s is driven in the opposite direction at 35 m/s when it is hit by a bat. The average force on the ball is 6400 N. How long was it in contact with the bat? ( Impulse = change in momentum. ) F ∆ t = mvf – mvo = (0.2 kg)(35 m/s) – (0.2 kg)(-20 m/s) (6400 N) ∆ t = 11 kg m/s;
∆t 35 m/s

+
-20 m/s

∆ t = 1.72 ms

*9-6. A bat exerts an average force of 248 lb on 0.6-lb ball for 0.01 s. The incoming velocity of the ball was 44 ft/s. If it leaves in the opposite direction what is its velocity? Choose positive + as direction away from the bat, making incoming ball velocity negative: F ∆ t = mvf – mvo ; F ∆ t = mvf – mvo ; m= 0.6 lb = 0.01875 slugs 32 ft/s 2

(240 lb)(0.01 s) = (0.01875 slugs)vf - (0.01875 slugs)(-44 ft/s) 0.01875 vf = 2.4 lb s – 0.825; vf = 84.0 ft/s

*9-7. A 500-g ball travels from left to right at 20 m/s. A bat drives the ball in the opposite direction with a velocity of 36 m/s. The time of contact was 0.003 s. What was the average force on the ball? ( m = 0.5 kg, vo = +20 m/s, vf = -36 m/s, ∆ t = 0.003 s )

F ∆ t = mvf – mvo ; F(0.003 s) = (0.5 kg)(-36 m/s) – (0.5 kg)(20 m/s) F= −18 kg m/s - 10 kg m/s ; 0.003 s F = 9330 N

*9-8. A 400-g rubber ball is dropped a vertical distance of 12 m onto the pavement. It is in contact with the pavement for 0.01 s and rebounds to a height of 10 m. What is the total change in momentum? What average force is exerted on the ball?

102

01 s) = (0.0 m/s px = (0.80 m/s 9-10. If the mass of the ball is 200 g.5 ms.99 kg m/s.3 m/s . What are the x.0015 s 1. A golfer hits a 46 g golf ball with an initial velocity of 50 m/s at 300.15 kg m/s. 6th Edition To apply the impulse-momentum theorem.15 kg m/s .3 m/s vf = + 14 m/s v f = 2(9. (80 N)(0. The face of the club in Problem 9-10 is in contact with the ball for 1.2 kg)vf – 0.0015 s Fx = 1330 N Fy = Fy = 767 N 103 . (Ep)Beginning = (Ek)Ground .99 kg m/s.Chapter 9 Impulse and Momentum Physics. Fx ∆ t = pfx – pox =1. *9-9. What are the horizontal and vertical components of the average force on the ball? We need to treat horizontal and vertical impulses and momenta separately: From previous problem: po = 0.8 m/s 2 )(12 m) ½mvf2 = mghf .15. 0.99 kg m/s. py = 1. A cue stick strikes an eight-ball with an average force of 80 N over a time of 12 ms.99 kg m/s . vo = . what will be its velocity? F∆ t = mvf – mvo.3 m/s). pf = 1. v = 4.046 kg)(25 m/s) vy 300 50 m/s vx px = 1.4 kg)(14 m/s) – (0. vy = (50) sin 300 = 25.046 kg)(43. Fx = 1. py = (0. 0. 12 m vf hf 10 m vo F = 1170 N v0 = 2 gh0 = 2(9.4 kg)(-15. mgho = ½mvo2.8 m/s 2 )(10 m) F∆ t = mvf – mvo.3 m/s). pfy = 1. we need to first find the velocities just before and just after impact with the ground.and y-components of the momentum imparted to the ball? vx = (50) cos 300 = 43.15 m/s *9-11.012 s) = (0. F(0.15 kg m/s Fx ∆ t = pfx – pox =1.

The child jumps forward at 2 m/s. v1 = −m2 v2 −(2 kg)(9 m/s) = . v2 (-12 m/s) m2 = 1.00 m/s 9-13.Chapter 9 Impulse and Momentum Physics. what was the mass of the football? Momentum zero before and after: 0 + 0 = m1v1 + m2v2 m2 = −m1v1 −(70 kg)(0. A spring is tightly compressed between a 6-kg block and a 2–kg block and then tied with a string. The connecting string breaks and sends the smaller mass to the left with a velocity of 10 m/s. 9-8.3. are compressed against a spring and then tied together on a frictionless surface as shown in Fig. What is the velocity of the 6-kg block? kg v2 v1 Total momentum is zero before and after the event. If the person moves backward at 34 cm/s. When the string breaks. 6th Edition Conservation of Momentum 9-12. 0 + 0 = m1v1 + m2v2 . Two masses. A 20-kg child is at rest in a wagon.3.98 m/s 9-15. A 70-kg person standing on a frictionless surface throws a football forward with a velocity of 12 m/s. m1 (6 kg) v1 = .34 m/s) = . m2 (3m) v1 = . one three times that of the other. What is the mass of the wagon? 104 . What was the velocity of the larger mass? Momentum zero before and after: 0 + 0 = m1v1 + m2v2 v2 = −m1v1 −m(10 m/s) = . sending the wagon backward at 12 m/s. the 2-kg block moves to the right with a velocity of 9 m1 = 6 kg m2 = 2 kg m/s.33 m/s 10 m/s m v1 3m v2 9-14.

3 m/s m1 (45 g) *9-18. 6th Edition 0 = m1v1 + m2v2 . Find the recoil velocity of the rifle and the ratio of the kinetic energy of the bullet to that of the rifle? 0 = m1v1 + m2v2 .4 m/s. weighing 80 and 50 lb.50 m/s 105 .13. v1 = Thus.6 m/s + 5. v1 = −m2 v2 −(50 lb)(6 ft/s) = .33 kg v1 (-12 m/s) Physics. A 6-kg bowling ball collides head on with 1. sending a 45-g piece to the left and another to the right with a velocity of 40 m/s.3.8 kg)(3 m/s) 6ub = 9.Chapter 9 Impulse and Momentum m1 = − m2 v2 −(20 kg)(2 m/s) = .6 m/s) + (1.32 m/s) 2 Ratio = 208 *9-19.8-kg pin.75 ft/s m1 (80 lb) (Here were able to use the weight because it is proportional to the mass) 9-17. (6 kg)ub = (6 kg)(1.32 m/s 2 Ekb ½ mb vb (24 g)(900 m/s) 2 = = . are at rest on roller skates. 9-16. v1 = . What was the initial velocity of the bowling ball? mbub + 0 = mbvb + mpvp. m1 (5000 g) v1 = . The larger child pushes so that the smaller moves away at 6 mi/h. v1 = . m1 = 45 g.4.3. 0 = m1v1 + m2v2 . m1 = . ub = 2. Two children. A 60-g firecracker explodes. v1 = −m2 v2 −(24 g)(900 m/s) = . A 24-g bullet is fired with a muzzle velocity of 900 m/s from a 5-kg rifle. The pin moves forward at 3 m/s and the ball slows to 1. What is the velocity of the larger child? 0 = m1v1 + m2v2 . Ekr ½ mr vr2 (5000 g)(4. m2 = 15 g −m2 v2 −(15 g)(40 m/s) = .6 m/s. What is the velocity of the 45-g piece? The two pieces add to 60 g: m1 + m2 = 60 g.

0 J – 0.96 m/s.625 J. assuming the cars stick together? ( u1 = 8.625 m/s)2 + Loss Loss =10.48 m/s 2ub = 4. (0. u2 = 0. What is the velocity of the wreckage immediately after the crash.15 J – 0.198 J. ½mrur2 + 0 =(mr + mb)vc2.2 kg)(0.2 kg)(10 m/s)2 = ½ (3.625 m/s 2 m/s = 3. (a) If the two stick together on collision. A car traveling at 8 m/s crashes into a car of identical mass stopped at a traffic light. What was the velocity of the ball before it was caught? What energy was lost in the process? (A completely inelastic collision: vc = vm = vb = 8 cm/s) mbub + mmum = (mb + mm)vc .08 m/s)2 + Loss Loss = 6.2 kg + 3 kg)vc vc = 0. The ball and man each move at 8 cm/s after the ball is caught.95 J *9-21. ½(0. mu1 = 2mvc 106 . Loss = 5.2 kg)(10 m/s) + 0 = (0. (2 kg)ub + 0 = (2 kg + 60 kg)(0. 6th Edition *9-20. m1 = m2 = m) 0 mu1 + mu2 = (m + m)vc . A 200-g rock traveling south at 10 m/s strikes a 3-kg block initially at rest.38 J Elastic and Inelastic Collisions 9-22.00 m/s. Loss = 9. ½mbub2 + 0 =(mb + mm)vc2.Chapter 9 Impulse and Momentum Physics. ½(2 kg)(2. A 60-kg man on a lake of ice catches a 2-kg ball.48 m/s)2 = ½(62 kg)(0.08 m/s) ub = 2.2 vc . what will be their common velocity? (b) What energy was lost in the collision? mrur + mbub = (mr + mb)vc .

What velocity will the child have after catching the football? m1u1 + 0 = m1v1 + m2v2. (2000 kg)(10 m/s) + 0 = (2000 kg + 1200 kg)vc vc = 6.500 J 9-24. (20 g)(-8 m/s) + (10 g)(5 m/s) = (20 g + 10 g)vc vc = -3. Loss = 37. vc =0. to left -110 m/s = 30 vc .8 kg)(15 m/s) = (30 kg + 0. *9-26. %Loss = 563 J = 73. The father throws a 0. 6th Edition vc = 4. Find the percent loss of energy for the collision in Problem 9-25.25 m/s 20. (0.000 m/s = 3200 vc . A 2000-kg truck traveling at 10 m/s crashes into a 1200-kg car initially at rest. A 20-g object traveling to the left at 8 m/s collides head on with a 10-g object traveling to the right at 5 m/s. 2 2 Physics.500 J.8 kg)vc = 12 m/s.8-kg football with a velocity of 15 m/s. What is their combined velocity after impact? m1u1 + m2u2 = (m1 + m2)vc . ½(2000 kg)(10 m/s)2 = ½(3200 kg)(6.Chapter 9 Impulse and Momentum vc = u1 8 m/s = . A 30-kg child stands on a frictionless surface. ½m1u12 + 0 =(m1 + m2)vc2. Loss = 563 J.6% 765 J 107 . What is the common velocity after the collision if they stick together? What is the loss in energy? m1u1 + m2u2 = (m1 + m2)vc .390 m/s *9-25.00 m/s 9-23.000 J – 62.8 kg)vc (30.67 m/s)2 + Loss 765 J = 202 J + Loss.67 m/s. Conservation of Energy: ½m1u12 + ½m2u22 =½(m1 + m2)vc2 + Loss ½(20 g)(-8 m/s)2 + ½(10 g)(5 m/s)2 = ½(30 g)(-3.25 m/s)2 + Loss Loss = 100.

8 m/s 2 )(0.5 kg + 2 kg)(1.98 m/s (0.20 m) . m1u1 + 0 = (m1 + m2) vc .9 m/s *9-29.09 m) . we need to know the common velocity of the clay and ball after the collision.20 m) .5 kg)u1 = (0. v2 = 1.8 m/s 2 )(0. What was the initial velocity of the bullet if the combined masses rise to a height of 9 cm? ½(m1 + m2) vc2 = (m1 + m2) gh vc = 2 gh = 2(9. vc = 1.95 m/s. A 2-kg block of clay is tied to the end of a string as shown in Fig. h 10 m/s Momentum is conserved: m1u1 + 0 = m1v1 + m2v2.98 m/s) u1 = 9. what must be the new entrance velocity if the block is to raise to the same height of 20 cm? We must find the velocity v2 of the clay (m2) after collision: ½(m1 + m2) v22 = (m1 + m2) gh vc = 2 gh = 2(9.33 m/s) 108 . Find the entrance velocity of the ball? h Before applying momentum conservation.5 kg) u1 = (0.009 kg) u1 = (0. A 9-g bullet is embedded into a 2.90 m/s (0. Energy is conserved : ½(m1 + m2) vc2 = (m1 + m2) gh vc = 2 gh = 2(9. vc = 1.33 m/s (0.98 m/s). 9-9.5 kg)u1 = 4.Chapter 9 Impulse and Momentum Physics.5 kg)(10 m/s) + (2 kg)(1. In Problem 9-27. A 500-g steel ball embeds itself into the clay causing both to rise a height of 20 cm.8 m/s 2 )(0. *9-28. (0.009 kg + 2 kg)(1.98 m/s. u1 = 17. 6th Edition *9-27. suppose the 500-g ball passes entirely through the clay an emerges with a velocity of 10 m/s.0 kg ballistic pendulum (see Fig. 8-13). m1u1 + 0 = (m1 + m2) vc .

Substituting this for v2.07 s.68 m/s. v 0 = 2 gh = 2(9. 9-31. h2 =5. u1 = 297 m/s Physics. T = 2t. A billiard ball moving to the left at 30 cm/s collides head on with another ball moving to the right at 20 cm/s.67 m) = . v2 = 0 m(-30 cm/s) + 0 = mv1 + mv2 .54 m/s 2 ½ mv0 = mgh.50 cm/s From second equation: v2 = v1 – 50 cm/s.50 cm/s) = . The masses of the balls are identical.Chapter 9 Impulse and Momentum (0. v1 = -30 cm/s. t = 2s 2(5.9) 2 .67 m) . v1 + v2 = (-30 cm/s) + (20 cm/s). we obtain: v1 + (v1 . to left And. h1 h2 = h1e 2 = (7 m)(0.8 m/s 2 )(5. how high will it rebound? e= h2 . and v1 = 20 cm/s.67 m. v2 – v1 = . v1 + v2 = -10 cm/s Energy (e = 1): v2 – v1 = u1 – u2 = (-30 cm/s) – (20 cm/s). h1 e2 = h2 .10 cm/s.) Momentum: m1u1 + m2u2 = m1v1 + m2v2 Given: m1 = m2 = m. 6th Edition *9-30.009 kg)u1 = 2.90. then find t.54 m/s T = 2. If a steel ball is dropped from a height of 7 m. The coefficient of restitution for steel is 0. to right v2 = -30 cm/s. vo + 0 10. s= v0 + v f 2 t. v2 = v1 – 50 cm/s = (20 cm/s) – 50 cm/s. 109 .67 m h1 7m h2 *9-32. What is the time between the first contact with the surface and the second contact for Problem 9-31? (We need to know vo to rise to 5. what is the velocity of each ball after impact? (Consider right as +.15 s t = 1. If the collision is perfectly elastic.) vo = 10.

3 kg + 0.2 kg)(-3. v1 = 2 gh = 2(9. m2 = 200 g.100 cm/s (300 g)(50 cm/s) + (200 g)(-100 cm/s) = (300 g)v1 + (200 g)v2 110 . we obtain: Loss = 0. south (Note: When working with energy. 9-34 is perfectly elastic. u1 = 50 cm/s. h1 v2 = u2 = 0. vc = -10 cm/s. e = 0. A 300-g block moving north at 50 cm/s collides with a 200-g block moving south at 100 cm/s.11v1. u1 h1 −v1 −v1 = . v1 = -12.11v1 e= e= u1 = ½ mv12 = mgh. u1 − u2 h2 = 0. u1 = -1.8 m/s 2 )(8 m) . Suppose the collision in Prob. 6th Edition *9-33.135 J *9-35.67 m/s)2 + Loss Solving for “loss”.2 kg)(5 m/s)2 = ½(0. m1u1 + m2u2 = m1v1 + m2v2. u2 = . A ball dropped from rest onto a fixed horizontal plate rebounds to a height that is 81 percent of its original height. what is their common velocity after sticking together? What is the loss in energy? (Consider north as positive) Momentum: m1u1 + m2u2 = m1v1 + m2v2 . e (0. u1 = -1.000 g cm/s – 20.9) + h2 u1 = -1. What is the coefficient of restitution? What is the required velocity on the first impact to cause the ball to rebound to a height of 8 m.9 m/s *9-34. e= v2 − v1 .5 m/s u1 = 13. If the collision is completely inelastic.) Conservation of Energy: ½m1u12 + ½m2u22 =½(m1 + m2)vc2 + Loss ½(0.Chapter 9 Impulse and Momentum Physics.3 kg)(-8 m/s)2 + ½(0. m1 = 300 g.900 −v1 . v1 = v2 = vc for inelastic collision (300 g)(50 cm/s) + (200 g)(-100 cm/s) = (300 g + 200 g)vc 15.000 g cm/s = (500 g)vc.5 m/s). it is necessary to use kg for the mass unit.11(-12. Find the velocities after impact.81.

111 ∆ t = 3. What will be their velocities after impact if the collision is (a) completely inelastic or (b) perfectly elastic? Since weight is proportional to mass. we will use the weights instead. 6th Edition Energy (e = 1): v2 – v1 = u1 – u2 = (50 cm/s) – (-100 cm/s).00 ms .4 kg)(30 m/s) – 0.50 cm/s. v2 – (-80 cm/s) = 150 cm/s. A 5-lb and a 12-lb object approach each other with equal velocities of 25 ft/s. Elastic case: (5 lb)(25 ft/s) + (12 lb)(-25 ft/s) = (5 lb)v1 + (12 lb)v2 .Chapter 9 Impulse and Momentum Dividing each term by 100 g: 3 v1 + 2 v2 = -50 cm/s Physics. v2 – v1 = 50 ft/s Substitute v1 = v2 . to left v2 = 70 cm/s.41 ft/s v1 = -45.50 ft/s into the earlier equation and solve for v1: (v2 – 50 ft/s) + 2.35 ft/s. to right *9-36. v1 = -80 cm/s. Momentum: W1u1 + W2u2 = W1v1 + W2v2 .3 ft/s (5 lb)(25 ft/s) + (12 lb)(-25 ft/s) = (5 lb + 12 lb)vc.6 ft/s Challenge Problems 9-37.4 v2 = -35 ft/s Energy (e = 1): v2 – v1 = u1 – u2 = (25 ft/s) – (–25 ft/s). What was the time of contact for this force? F ∆ t = mvf – mvo = (0. v2 = 4. Dividing each term by 5 lb: v1 + 2. v1 = v2 – 50 ft/s = 4.4 v2 = .41 ft/s – 50 ft/s. v2 – v1 = 150 cm/s Substitute v2 = v1 + 150 cm/s into the earlier equation and solve for v1: 3 v1 + 2 (v1 + 150 cm/s) = . An average force of 4000 N acts on a 400-g piece of metal causing it to move from rest to a velocity of 30 m/s. v1 = v2 = vc for inelastic collision vc = -10. (4000 N)∆ t = 12 kg m/s.

02 kg)(10 m/s) + (10 kg)v2 . The coefficient of restitution is 0.2 kg)(10 m/s)2 + ½(10 kg)(0.5 v2 = 25 cm/s v2 – v1 = e(u1 – u2).02 kg (0. we obtain: Loss =3990 J v2 = 0. 2 2 v f = −8. m1 = 60 g.380 m/s v2 = ? v1 = 10 m/s *9-41. . What impulse was applied by the wall? ½mv02 = 2(½mvf2) .6 kg)(12 m/s).6 kg)( –2.3 N m *9-39.2 kg)(200 m/s)2 = ½(0.8. 9-40. A 10-kg block at rest on a horizontal surface is struck by a 20-g bullet moving at 200 m/s. u2 = .380 m/s)2 + Loss Solving for “loss”. 6th Edition *9-38.30 cm/s (60 g)(100 cm/s) + (150 g)(-30 cm/s) = (60 g)v1 + (150 g)v2 Divide each term by 60 g and simplify: v1 + 2. 2 v0 (12 m/s) 2 vf = = .Chapter 9 Impulse and Momentum Physics. u1 = 100 cm/s. v2 – v1 = 104 cm/s Solve for v1: v1 = v2 – 104 cm/s. In Problem 9-39. What are the velocities after impact. how much kinetic energy was lost? Conservation of Energy: ½m1u12 + 0 = m1v12 + m2v22 + Loss ½(0. An 600-g object whose velocity is initially 12 m/s. The bullet passes entirely through the block and exits with a velocity of 10 m/s.8[100 cm/s – (-30 cm/s)]. F ∆ t = -12. collides with a wall and rebounds with half of its original kinetic energy. 112 Now substitute to find v2. A 60-g body having an initial velocity of 100 cm/s to the right collides with a 150-g body moving to the left at 30 cm/s. v2 – v1 = 0. What percent of the energy is lost in collision? m1u1 + m2u2 = m1v1 + m2v2. What is the velocity of the block? m1u1 + 0 = m1v1 + m2v2.49 m/s F ∆ t = mvf – mvo = (0.45 m/s) – (0. m2 = 150 g. m1 = 0.02 kg)(200 m/s) = (0.

what is their common velocity? m1 = m2 = m3 = m. to right v1 = -67.) 10 m/s – 2 m/s – 2 m/s = 3 vc.15 kg)(-0. 2g 2(32 ft/s 2 ) h = 0. (0.625 ft/s Now.Chapter 9 Impulse and Momentum *9-41.369 m/s)2.93 in.03675 J Efk = ½(0. If all three couple together after colliding. Physics. u1 = 10 m/s.) (v2 – 104 cm/s) + 2.4 kg + 6 lb)vc m∝W h vc = 5.494 ft.4 lb)(90 ft/s2) = (0.00 m/s. Conservation of Energy: ½m1u12 + ½m2u22 = m1v12 + m2v22 + Loss For energy we must use SI units with mass in “kg” and velocity in ”m/s. Efk = 0. A single railroad car moving north at 10 m/s strikes two identical. The block in Fig.06 kg)(1 m/s)2 + ½(0. vc = 2. How high will it rise if a 0.5% *9-42.1 cm/s. (Cont. *9-43.671 m/s)2 + ½(0.06 kg)(-0.625 ft/s) 2 h= = . u2 = u3 = -2 m/s. 6th Edition v2 = 36.15 kg)(0.03675 J    %Loss = 35.0237 J   = 100   0. to left v1 = v2 – 104 cm/s = (36.4-lb projectile with an initial velocity of 90 ft/s embeds itself into the block? Momentum is conserved: m1u1 + 0 = m1v1 + m2v2. 9-6 weighs 6 lb.9 cm/s. 2 v0 (5.0.0237 J  E − E fk % Loss = 100  ok  Eok   0.5 v2 = 25 cm/s.03675 J . h = 5. north 113 . v1 = v2 = v3 =vc Momentum is conserved: mu1 + m(u2 + u3) = (3m)vc (Mass divides out. coupled cars initially moving south at 2 m/s.9 cm/s) – 104 cm/s. The mass divides out. we can find h using conservation of energy and the common initial velocity vc: ½(m1 + m2) v22 = (m1 + m2) gh .” Eok = ½(0. Eok = 0.3 m/s)2.

6 v2 = 4 x 106 m/s Energy (e = 1): v2 – v1 = u1 – u2 = 4 x 106 m/s – 0 . u1 = 4 x 106 m/s (2 x 10-27 kg)( 4 x 106 m/s) = (2 x 10-27 kg)v1 + (1. 6th Edition *9-44.0 m/s) – (0. v2y = (60 sin 300) = 30 m/s Fx ∆ t = J = mv2x – mv1x . Jy = (0. v1y = 0 v2x = (60 cos 300) = 52. It leaves the bat with a velocity of 60 m/s at an angle of 300 with the horizontal. Fy ∆ t = J = mv2y – mv1y . Jy = 12.00 x 10-27 kg moves with a velocity of 4. Assuming a perfectly elastic collision. v1 = 1.0 N s v2y 300 60 m/s v1x v2x -20 m/s Jx = 28.0 N s = 2400 N 0.00 x 106 m/s *9-45.8 N s *9-46.00 x 106 and collides head on with a particle of mass 1.8 N s = 5760 N . If the bat in Problem 9-45 was in contact with the ball for 5 ms. m2 = 1. A bat strikes a 400-g softball moving horizontally to the left at 20 m/s. m1 = 2 x 10-27 kg. v2 – v1 = 4 x 106 m/s Substitute v2 = ( v1 + 4 x 106 m/s) into the earlier equation and solve for v1: v1 + 0.2 x 10-27 kg. An atomic particle of mass 2.4 kg)(30 m/s) – 0 .0 m/s. What are the horizontal and vertical components of the impulse imparted to the ball? First find components of velocity: v1x = .20 m/s.005 s F = 6240 N F = Fx2 + Fy2 = (5760 N) 2 + (2400 N) 2 .4 kg)(-20 m/s).Chapter 9 Impulse and Momentum Physics. what is the velocity of the incident particle after the collision? m1u1 + 0 = m1v1 + m2v2. 114 .005 s Fy = 12. what was the magnitude of the average force on the softball? Fx = 28. 0.2 x 10-27 kg )v2 Dividing each term by 2 x 10-27 kg: v1 + 0.4 kg)(52. Jx = (0.20 x 10-27 kg initially at rest.6 (v1 + 4 x 106 m/s) = 4 x 106 m/s.

620 m/s + 1. v1 = 15 m/s – 2(10 m/s ) = -5 m/s and v2 = 10 m/s v1 = -5 m/s 115 . What is the velocity of each mass after the collision? Elastic: m1u1 + 0 = m1v1 + m2v2 and v2 – v 1 = u1 – u2 (1 kg)(15 m/s) = (1 kg)v1 + (2 kg)v2 . v2 = 15 m/s + v1 v2 = 15 m/s + (15 m/s – 2v2). m2 = 234 g m/s 2.00 m/s Ratio = 0. Ek 1 ½ m1u12 (1 kg)(15 m/s) 2 *9-49. v1 = 15 m/s – 2 v2 v2 – v 1 = u1 – u2 = (15 m/s) – 0.02 m/s *9-48. m2 v2 = 234 g m/s. If the collision in Fig. v1 = 0. Assume the collision in Problem 9-48 is perfectly elastic.02 m/s m2 = 116 g v2 = v1 + 1. The collision is perfectly elastic and the 300-g cart’s velocity is reduced to 0.4 m/s when it hits a second cart B at rest. assume that the collision of the two masses is completely inelastic. A cart A has a mass of 300 g and moves on a frictionless air track at 1.Chapter 9 Impulse and Momentum Physics.4 m/s.4 m/s.620 m/s) + m2v2 m2v2 = 234 g m/s. v2 = 10 m/s .4 m/s = 0. Elastic collision: v2 – v1 = u1 – u2 = (1.333 1 kg 15 m/s 2 kg Ek 2 ½(m1 + m2 )vc2 (3 kg)(5 m/s) 2 = = . 9-10. v1 + 2v2 = 15 m/s .620 m/s m1u1 + 0 = m1v1 + m2v2. u1 = 15 m/s (1 kg)(15 m/s) + 0 = (1 kg + 2 kg)vc vc = 5. (300 g)(1. What is the common velocity after the collision and what is the ratio of the final kinetic energy to the initial kinetic energy? m1u1 + 0 = (m1 + m2)vc .4 m/s) = (300 g)(0.620 m/s after the collision. 6th Edition *9-47. u1 = 1. What was the mass of the other cart and what was its velocity after the collision? m1 = 300 g.4 m/s) – 0 v2 = 2.

u2 = . v2 – v1 = 6 m/s Substitute v2 = v1 + 6 m/s into the earlier equation and solve for v1: v1 + 3(v1 + 6 m/s) = .00 m/s 116 .0 J **9-51. (a) If the collision is completely inelastic. we obtain: Loss = 27. v2 – (-7.50 m/s)2 + Loss Solving for “loss”.4 m/s (2 kg)(2 m/s) + (6 kg)(-4 m/s) = (2 kg)v1 + 6 kg)v2 Dividing each term by 2 kg: v1 + 3 v2 = -10 m/s Energy (e = 1): v2 – v1 = u1 – u2 = (2 m/s) – (-4 m/s). vc = -2.00 m/s) = 6 m/s. m2 = 6 kg. A 2-kg mass moves to right at 2 m/s and collides with a 6-kg mass moving to the left at 4 m/s. 6th Edition *9-50.Chapter 9 Impulse and Momentum Physics. m1 = 2 kg. u1 = 2 m/s. assume the collision is perfectly elastic. What are the velocities after the collision? m1u1 + m2u2 = m1v1 + m2v2.10 m/s. v1 = v2 = vc for inelastic collision (2 kg)(2 m/s) + (6 kg)(-4 m/s) = (2 kg + 6 kg)vc 4 kg m/s – 24 kg m/s = (8 kg)vc. what is their common velocity after colliding. In Problem 9-50.50 m/s Conservation of Energy: ½m1u12 + ½m2u22 = ½(m1 + m2)vc2 + Loss ½(2 kg)(2 m/s)2 + ½(6 kg)(-4 m/s)2 = ½(2 kg + 6 kg)(-2. v1 = -7. and how much energy is lost in the collision? Momentum: m1u1 + m2u2 = m1v1 + m2v2 .00 m/s v2 = -1.

could we use the weight of the objects instead of the mass? Why.675 m/s: vf = 10(0. vf = .25 slugs 2 32 ft/s 32 ft/s 2 Ekb = 10 (½mvb2) = (5)(0.75 ft/s)2 . va = -0. the weight can be used instead of the mass to calculate either velocities or weights of colliding objects.05 lb)(2700 ft/s) = . In applying conservation of momentum for colliding objects to find final velocities.00156 slugs .6. Since weight is proportional to mass: W = mg. what is the final velocity of the astronaut? Compare the final kinetic energy of the ten bullets with that of the astronaut. An astronaut in orbit outside a capsule uses a revolver to control motion.00156 slugs)(2700 ft/s)2. va = −Wb vb −(0.950 ft lb Eka = ½mava2 = ½(6. mb = = 6. The speeds dominate. 9-36.25 slugs)(6.75 ft/s We need masses: mb = 0. and 10 shots are fired. 6th Edition Critical Thinking Questions *9-52. For example. m1u1 + m2u2 = m1v1 + m2v2 W1 W W W u1 + 2 u2 = 1 v1 + 2 v2 g g g g W1u1 + W2u2 = W1v1 + W2v2 117 .05-lb bullets at 2700 ft/s.675 m/s). see Prob. or why not? Verify your answer by applying it to one of the examples in the text. If the revolver fires 0. Why is the difference so great? 0 + 0 = Wava + Wbvb. The astronaut with gear weighs 200 lb on the earth. Eka = 142 ft lb The kinetic energy of the bullets is much larger because when finding the kinetic energy. Ekb = 56.Chapter 9 Impulse and Momentum Physics.05 lb 200 lb = 0. *9-53. and since mass appears in every term involving conservation of momentum.675 ft/s Wa 200 lb Each shot changes va by –0. one must deal with the square of velocity.

02 kg)(0.020 kg)(200 m/s) = (0. The masses do not change in the collision.20 m/s. strikes a 10-kg wooden block and passes entirely through it. Fx = 5520 N Fy ∆ t = mv2y – mv1y = (0. so the change in momentum of one should be the negative of the change for the other. v2 = 0. v1y = 0 v2x = (60 cos 300) = 52. Is this true whether the collision is elastic or inelastic.3 kg)(-40 m/s). equal but opposite impulses should always produce equal and opposite changes in momentum. The ball is in contact with the bat for a time of 5 ms. emerging with a velocity of 10 m/s.005 s) = 27. Fy(0. A 0.00 N s . what are the horizontal and vertical components of the average force acting on the bat? First find components of velocity: v1x = . Fy = 1800 N v2y 300 60 m/s v1x v2x -40 m/s *9-56. and it leaves with a velocity of 60 m/s at an angle of 300. When two masses collide they produce equal but opposite impulses.3 kg)(30 m/s) – 0 .0 m/s) – (0.38 m/s)2 + Loss Loss = 399 J *9-55.5 m/s 118 .3 kg)(52. 9-50: The test is whether: m1v1 – m1u1 = -( m2u2 – m2v2).02 kg)(200 m/s)2 = ½(10. 6th Edition *9-54. Prob.0 m/s.30-kg baseball moving horizontally at 40 m/s is struck by a bat.6 N s .Chapter 9 Impulse and Momentum Physics. Verify your answer by using the data in Problems 9-50 and 9-51. Momentum is conserved whether energy is lost in collision or not.020 kg)(10 m/s) + (10 kg)v2 . moving at 200 m/s. What is the velocity of the block after impact? How much energy was lost? m1u1 + 0 = m1v1 + m2v2 (0. A 20-g bullet.380 m/s Conservation of Energy: ½m1u12 + 0 = ½(m1 + m2)vc2 + Loss ½(0. v1 = v2 = -2. v2y = (60 sin 300) = 30 m/s Fx ∆ t = mv2x – mv1x = (0. Fx(0.005 s) = 9. Therefore.

After the collisions. u1 = 5 m/s. It works for inelastic collisions.18 kg m/s = . m2 = 3m. 6th Edition (2 kg)(-2. The first block has a mass of 1-kg and the second has a mass of 2-kg. v1 = -7 m/s.5 m/s ) m1u1 + m2u2 = (m1 + m2 ) vc . the 1kg block moves with a velocity of 1 m/s and the 2-kg block moves with 2 m/s. v2 = -1 m/s (2 kg)(-7 m/s) . u2 = . The bullet passes completely through the first block and lodges into the second.9 kg m/s. An 8-g bullet is fired horizontally at two blocks resting on a frictionless surface. v1 + 3v2 = -10 m/s.(2 kg)(2 m/s) = -[(6 kg )(-2. *9-57. Two toy cars of masses m and 3m approach each other.(6 kg)(-4 m/s)] .10 m/s. m(5 m/s) + 3m(-5 m/s) = mv1 + 3mv2 Now for elastic: v2 – v1 = u1 – u2 v1 = v2 – 10 m/s v2 = 0 v2 – v1 = 5 m/s – (–5 m/s) = 10 m/s.50 m/s) . (Cont.50 m/s) . It also works for elastic collisions. -10 m/s = 4 vc. What is the velocity of the bullet before and after emerging from the first block? --Observe the figure on the next page -- 119 . 9-51: Same test: m1v1 – m1u1 = -( m2v2 – m2u2).9 kg m/s = .(2 kg)(2 m/s) = -[(6 kg )(-1 m/s) . m(5 m/s) + 3m(-5 m/s) = (m + 3m) vc For inelastic case vc = -2. v1 = -10 m/s *9-58. Prob. v1 = (0) – 10 m/s = . (v2 – 10 m/s) + 3 v2 = -10 m/s.(6 kg)(-4 m/s)] . each traveling at 5 m/s. If they couple together.50 m/s m1u1 + m2u2 = m1v1 + m2v2 .Chapter 9 Impulse and Momentum *9-56. what is their common speed afterward? What are the velocities of each car if the collision is perfectly elastic? ( m1 = m.18 kg m/s.) Physics.

mAuA + 0 = mAvA + mBvB. we apply conservation to first block only: (0. From which: vA = vB – 3. vB – vA = 3.96 m/s L = 80 cm mA = 1 kg.96 m/s.64 m/s (vB – 3.008 kg)ve + (1 kg)(1 m/s) . Assuming that the collision is perfectly elastic what are the velocities of each mass immediately after impact? First find uA from energy of fall: ½mv2 = mgh v = 2 gh = 2(9.96 m/s (1 kg)(3.008 kg) v1 = (1 kg)(1 m/s) + (2.96 m/s .32 m/s 120 . (Cont. v1 = 627 m/s To find velocity emerging from 1-kg mass. A 1-kg mass A is attached to a support by a cord of length 80 cm. 6th Edition 8g *9-58.96 m/s = 2.96 m/s) + 2 vB = 3.) 1 kg 1 m/s Total momentum at start = total momentum at finish (0. ve = 502 m/s *9-59.008 kg)(2 m/s).8 m) .96 m/s – 0 mA = 2 kg vA = vB – 3. vB = 2. 9-11.96 m/s . After release it swings downward striking the 2-kg mass B which is at rest on a frictionless tabletop.96 m/s .96 m/s) = (1 kg)vA + (2 kg) vB vA + 2 vB = 3. mB = 2 kg Elastic: vB – vA = uA – uB = 3.Chapter 9 Impulse and Momentum 2 kg 2 m/s Physics. vA = -1. Substitute for vA in the other equation.008 kg)(627 m/s) = (0.8 m/s 2 )(0. v = 3.64 m/s – 3. uA = 3.96 m/s. and it is held horizontally as in Fig.

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