Momentum Practice Questions
1. Find the momentum of a ...
A. 60.0 kg person moving east at 9.0 m/s.
B. 1.0 x 103 kg car moving north at 20.0 m/s.
C. 4 x 101 kg person moving south at 2.0 m/s.
2. A car has 2.0 x 105 units of momentum. What would be the car's new momentum if ...
a) its velocity were doubled
b) its velocity were tripled
c) its mass were doubled
d) both its velocity were doubled and its mass were doubled
3. A 0.50-kg cart (cart A) is pulled with a 1.0 N force for 1.0 s. Another 0.50 kg cart (cart B) is pulled with a 2.0 N-force for 0.50 s.
A. Which cart (A or B) has the greatest acceleration?
B. Which cart (A or B) has the greatest impulse?
C. Which cart (A or B) has the greatest change in momentum? Explain your answer.
4. A 0.105-kg hockey puck moving at 48 m/s is caught by a 75-kg goalie at rest. With what velocity does the goalie slide on the ice after catching the puck?
5. A 35.0 g bullet strikes a 5.0-kg stationary wooden block and embeds itself in the block. The block and bullet move together at 8.6 m/s. What was the original velocity of the bullet?
6. A 25.0 g bullet is initially moving at 475 m/s when it strikes a 2.5 kg wooden block. The bullet passes through the block, leaving at 275 m/s. The block was at rest when it was hit. How fast is the block moving when the bullet leaves?
7. A 0.50 kg ball traveling at 6.0 m/s collides head-on with a 1.00 kg ball moving in the opposite direction at a velocity of -12.0 m/s. The 0.50 kg ball moves away at -14.0 m/s after the collision. Find the velocity of the second ball.
8. A 3.0 x 103 kg truck moving rightward with a speed of 5.0 km/hr collides head-on with a 1.0 x 103 kg car moving leftward with a speed of 10.0 km/hr. The two vehicles stick together and move with the same velocity after the collision. Determine the post-collision velocity of the car and truck.
9. A 75.0 kg football player moving eastward with a speed of 8.0 m/s collides head-on with a 1.0 x 102 kg lineman moving westward with a speed of 4.0 m/s. The two players collide and stick together, moving at the same velocity after the collision. Determine the post-collision velocity of the two players.
10. A van has 5.00 times the mass of a small car. If the van coasts at 10.0 km/hr into the car that is initially at rest, how fast do the two coast if they attach together?
11. Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of 16.6 kg and an initial velocity of 7.13 m/s due east. Object B has a mass of 29.1 kg and an initial velocity of 5.28 m/s due north. Find:
a) magnitude of the velocity after the collision
b) direction of the total momentum of this two-object system after the collision.
c) the amount of kinetic energy lost after the collision
12. A 45 kg swimmer runs with a horizontal velocity of 5.1 m/s off a boat dock onto a stationary 12 kg raft.
a) find the velocity that the swimmer and raft have after the impact not considering any friction.
b) find the total initial kinetic energy
c) find the total final kinetic energy
d) determine the amount of kinetic energy lost
13. In a football game, a receiver is standing still, having just caught a pass. Before he can move, a tackler, running at a velocity of 4.5 m/s, grabs him. The tackler holds onto the receiver, and the two move off together with a velocity of 2.6 m/s. The mass of the tackler is 115 kg. Assuming momentum is conserved, find the mass of the receiver.
14. A 7.5 kg ball moving right at 8.0 m/s collides head-on with a 5.0 kg ball moving to the left at 10.0 m/s. Determine the direction and speed of each ball after this elastic collision.
15. A 23.0 kg ball traveling right at 3.50 m/s collides elastically with a 15.5 kg ball moving left at 20.7 m/s. Find the speed and direction of each ball after the collision.
16. A ball with a mass of 23 kg is moving east at a speed of 8.25 m/s while a second ball with mass 17 kg is moving due north at 6.38 m/s. They hit each other squarely and move as one. If momentum is conserved find the magnitude and direction of the momentum after the collision of this two-object system.
17. The diagram below shows two balls moving towards each other and about to collide elastically. Using the magnitudes of velocity and the angles below, determine the magnitude and direction of the momentum of m1 after the collision.
18. The diagram below shows two balls moving towards each other and about to collide elastically. Using the magnitudes of velocity and the angles below, determine the magnitude and direction of the momentum of m1 after the collision.
19. For a constant force, if the duration of impact on an object is tripled, how much is the impulse increased?
20. Distinguish between impulse and momentum. Which is force x time, and which is inertia in motion?
21. Explain why it is a good thing for a driver if the time of impact in a collision lasts longer.
22. Why does more impulse happen during a collision when bouncing occurs than during a collision when it doesn’t? (recall that impulse = ∆momentum, and that momentum is a vector quantity)
23. For any closed system, what does it mean to say that the system’s momentum is conserved?
24. What is the difference between an elastic and inelastic collision?
25. A force of 940.0 N acts on a bicycle for 8.5 seconds.
a) what is the impulse on the bicycle?
b) what change in momentum occurs on the bicycle?
c) what is the final speed of the bicycle if the mass of the bicycle is 77 kg and is initially at rest?
26. What two things does a change in an object’s momentum depend on?
(ie. What two things does a mass in motion rely on to actually be a mass in motion?)
Solutions
1.
A. p = mv = (60.0 kg)(9.0 m/s) = 5.4 x 102 kgm/s, east
B. p = mv = (1.0 x 103 kg)(20.0 m/s) = 2.0 x 104 kgm/s, north
C. p = mv = (4 x 101 kg)(2.0 m/s) = 8 x 101 kgm/s, south
2.
A. p = 4.0 x 105 units (doubling the velocity will double the momentum)
B. p = 6.0 x 105 units (tripling the velocity will triple the momentum)
C. p = 4.0 x 105 units (doubling the mass will double the momentum)
D. p = 8.0 x 105 units (doubling the velocity will double the momentum and doubling the mass will also double the momentum; the combined result is that the momentum is quadrupled)
3.
A. Cart B has the greatest acceleration. Remember that acceleration depends on force and mass. They each have the same mass, but cart B has the greater force.
B. The impulse is the same for each cart. Remember that impulse is force x time, and the impulse for each cart is 1 Ns.
cart A… (1 N)(1 s) = 1 Ns…..cart B…(2 N)(0.5 s) = 1 Ns
C. The momentum change is the same for each cart. Remember that momentum change equals the impulse (Ft = ∆mv) Since we found in question B. that each cart has the same impulse, then each cart must also have the same momentum change.
4.
m1v1i + m2v2i = m1v1f + m2v2f
(0.105 kg)(48 m/s) + (75 kg)(0 m/s) = (0.105 kg)(vf) + (75 kg)(vf)
5.04 = 75.105(vf)
vf = 0.067 m/s, or 6.7 x 10-2 m/s
5. * 35 g = 0.035 kg
m1v1i + m2v2i = m1v1f + m2v2f
(0.035 kg)(vi) + (5.0 kg)(0 m/s) = (0.035 kg)(8.6 m/s) + (5.0 kg)(8.6)
(0.035)(vi) = 43.301
vi = 1237 or 1.2 x 103 m/s
6.
m1v1i + m2v2i = m1v1f + m2v2f
(0.025 kg)(475 m/s) + (2.5 kg)(0 m/s) = (0.025 kg)(275 m/s) + (2.5 kg)(vf)
11.875 = 6.875 + (2.5)(vf)
5.0 = (2.5)(vf)
vf = 2.0 m/s
7.
m1v1i + m2v2i = m1v1f + m2v2f
(0.50 kg)(6.0 m/s) + (1.00 kg)(-12.0 m/s) = (0.50 kg)(-14.0 m/s) + (1.00 kg)(vf)
3.0 - 12.0 = -7.0 + (1.00)(vf)
-9.0 + 7.0 = (1.00 kg)(vf)
vf = -2.0 m/s
8.
m1v1i + m2v2i = m1v1f + m2v2f
(3.0 x 103 kg)( 5.0 km/hr) + (1.0 x 103 kg)(-10.0 km/hr) = (3.0 x 103 kg)(vf) + (1.0 x 103)(vf)
15000 - 10000 = = (3.0 x 103 kg)(vf) + (1.0 x 103)(vf)
5000 = (4.0 x 103)(vf)
vf = 1.3 km/hr
9.
m1v1i + m2v2i = m1v1f + m2v2f
(75.0 kg)(8.0 m/s) + (1.0 x 102 kg)(-4.0 m/s) = (75.0 kg)(vf) + (1.0 x 102 kg)(vf)
600 - 400 = (75.0 kg)(vf) + (1.0 x 102 kg)(vf)
200 = (175)(vf)
vf = 1.1 m/s
10.
m1v1i + m2v2i = m1v1f + m2v2f
(5.00 units)(10 km/hr) + (1.00 unit)(0 km/hr) = (5.00 units)(vf) + (1.00 unit)(vf)
50 = (5.00)(vf) + (1.00)(vf)
50 = (6.00)(vf)
vf = 8.33 km/hr
11.
a)
x component:
(16.6)(7.13) + (29.1)(0) = (16.6)(v) + (29.1)(v)
118.358 = 45.7v
vx = 2.5898 m/s
y component:
(16.6)(0) + (29.1)(5.28) = (16.6)(v) + (29.1)(v)
153.648 = 45.7v
vy = 3.3621 m/s
vr2 = vx2 + vy2
vr2 = (2.5898)2 + (3.3621)2
vr = 4.243910985
vr = 4.24 m/s
b) tanθ = opp/adj
tanθ = 3.3621/2.5898
θ = 52.4o north of east
c)
Object A:
KEi = ½mivi2
KEi = ½(16.6)(7.13)2
KEi = 421.94627 J
Object B:
KEi = ½mivi2
KEi = ½(29.1)(5.28)2
KEi = 405.63072 J
Total KEi = 421.94627 + 405.63072 = 827.57699 J
KEf = ½mfvf2
KEf = ½(16.6 + 29.1)( 4.243910985)2
KEf = ½(45.7)( 4.243910985)2
KEf = 411.5463333 J
ΔKE = KEi - KEf
ΔKE = 827.57699 - 411.5463333 = 416.0306567 J
ΔKE = 416 J
12.
a) m1v1i + m2v2i = m1v1f + m2v2f
(45)(5.1) + (12)(0) = (45)(vf) + (12)(vf)
229.5 = (57)(vf)
vf = 4.0263 = 4.0 m/s
b)
KEi = ½mivi2
KEi = ½(45)(5.1)2
KEi = 585.225 = 5.9 x 102 J
c)
KEf = ½(45 + 12)(4.0263)2
KEf = 462.016 = 4.6 x 102 J
d) 585.225 - 462.016 = 123.209 = 1.2 x 102 J
13.
m1v1i + m2v2i = m1v1f + m2v2f
(m1)(0) + (115)(4.5) = (m1)(2.6) + (115)(2.6)
517.5 = (m1)(2.6) + 299.0
218.5 = (m1)(2.6)
m1 = 84.038 = 84 kg
14.
v1f = (m1 - m2)(v1i)/(m1 + m2)
v1f = (7.5 - 5.0)(18.0)/(7.5 + 5.0)
v1f = 3.6 m/s
v2f = (2m1)(v1i)/(m1 + m2)
v2f = (2)(7.5)(18.0)/(7.5 + 5.0)
v2f = 21.6 m/s
v1f = 3.6 -10.0 = -6.4 m/s
v2f = 21.6 - 10.0 = 11.6 m/s = 12 m/s
15.
v1f = (m1 - m2)(v1i)/(m1 + m2)
v1f = (23.0 - 15.5)(24.2)/(23.0 + 15.5)
v1f = (7.50)(24.2)/38.5
v1f = 4.714 = 4.71 m/s
v2f = (2m1)(v1i)/(m1 + m2)
v2f = (2)(23.0)(24.2)/(23.0 + 15.5)
v2f = 1113.2/38.5
v2f = 28.914 = 28.9 m/s
v1f = 4.714 - 20.7 = - 15.986 = -16.0 m/s
v2f = 28.914 - 20.7 = 8.214 = 8.21 m/s
16.
x component:
m1v1i + m2v2i = m1v1f + m2v2f
vxf = m1v1i + m2v2i
m1 + m2
vxf = (23 kg)(8.25 m/s) + (17 kg)(0.00 m/s)
(23 kg + 17 kg)
vxf = 189.75 kgm/s + 0.00 kgm/s
40 kg
vxf = 189.75 kgm/s
40 kg
vxf = 4.74375 m/s
y component:
m1v1i + m2v2i = m1v1f + m2v2f
m1v1i + m2v2i = vyf(m1 + m2)
vyf = m1v1i + m2v2i
m1 + m2
vyf = (23 kg)(0.00 m/s) + (17 kg)(6.38 m/s)
(23 kg + 17 kg)
vyf = 0.00 kgm/s + 108.46 kgm/s
40 kg
vyf = 108.46 kgm/s
40 kg
vyf = 2.7115 m/s
vr2 = vxf2 + vyf2
vr2 = (4.74375 m/s)2 + (2.7115 m/s)2
vr2 = 22.50316 m2/s2 + 7.3522322 m2/s2
vr2 = 29.855 m2/s2
vr = 5.5 m/s
tanθ = opp/adj
tanθ = 2.7115 /4.74375
θ = 3.0 x 101 o north of east
17.
pxi = pxf
pxi = p1xi + p2xi
pxi = m1v1xi + m2v2xi
pxi = (0.318)(2.83)(cos22.0o) + (0.215)(3.75)(cos15.0o)
pxi = 0.83440 + 0.77877
pxi = 1.61317
pxf = p1xf + p2xf
pxf = p1xf + (0.215)(2.27 m/s)(cos27.0o)
pxf = p1xf + 0.43485
pxi = pxf
1.61317= p1xf + 0.43485
p1xf = 1.61317 - 0.43485
p1xf = 1.17832
pyi = p1yi + p2yi
pyi = m1v1yi + m2v2yi
pyi = (0.318)(2.83)(sin22.0o) - (0.215)(3.75)(sin15.0o)
pyi = 0.337123 - 0.20867
pyi = 0.128453
pyf = p1yf + p2yf
pyf = p1yf + (0.215)(2.27 m/s)(sin27.0o)
pyf = p1yf + 0.22157
pyi = pyf
0.128453 = p1yf + 0.22157
p1yf = 0.128453 - 0.22157
p1yf = -0.093117
p1xf2 + p1yf2 = p1f2
p1f = 1.1819935 = 1.18 kgm/s
Ө = tan-1(p1yf/p1xf)
Ө = tan-1(-0.093117/1.17832)
Ө = -4.52o below the horizontal
p = mv
p1f = m1v1f
v1f = p1f/m1
v1f = (1.1819935)/(0.318)
v1f = 3.7169 = 3.72 m/s
18.
pxi = pxf
pxi = p1xi + p2xi
pxi = m1v1xi + m2v2xi
pxi = (18.0)(28)(cos20.0o) + (32.0)(25)(cos25.0o)
pxi = 473.60508 + 725.04622
pxi = 1198.65130
pxf = p1xf + p2xf
pxf = p1xf + (32.0 kg)(18 m/s)(cos37.0o)
pxf = p1xf + 460.01405
pxi = pxf
1198.65130 = p1xf + 460.01405
p1xf = 1198.65130 - 460.01405
p1xf = 738.63725
pyi = p1yi + p2yi
pyi = m1v1yi + m2v2yi
pyi = (18.0)(28)(sin20.0o) - (32.0)(25)(sin25.0o)
pyi = 172.37815 - 338.09460
pyi = -165.71645
pyf = p1yf + p2yf
pyf = p1yf + (32.0 kg)(18 m/s)(sin37.0o)
pyf = p1yf + 346.64545
pyi = pyf
-165.71645 = p1yf + 346.64545
p1yf = -165.71645 - 346.64545
p1yf = -512.36190
p1xf2 + p1yf2 = p1f2
p1f = 898.94365 = 9.0 x 102 kgm/s
Ө = tan-1(p1yf/p1xf)
Ө = tan-1(-512.36190/738.63725)
Ө = -35o or 35o below the horizontal
p = mv
p1f = m1v1f
v1f = p1f/m1
v1f = (898.94365)/(18.0)
v1f = 49.9413 = 5.0 x 101 m/s
19. Also tripled
20. Impulse is force x time, and momentum is inertia in motion.