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HON: HOMEWORK - IMPULSE-MOMENTUM & CONSERVATION OF
MOMENTUM - SOLUTIONS
9. m = 0.065 kg vi = 15 m/s, right vf = 15 m/s, left
F=?
F∆t = ∆p
F = ∆p/∆t
t = 0.020s
1. e. Momentum and impulse are measured in the same units
Ns = (kg m/s2)s = kg m/s
2. e. The bag increases the amount of time during which the
momentum is changing and reduces the average force on the
person.
3. W = mg m = W/g = 583 /9.80 = 59.5 kg
p = mv = (59.5 kg)( 1.63 m/s) = 97.0 kg m/s
left (final direction) is positive
Δp  0.065 (v  u)  0.06515 - - 15  1.95 kg m/s
F = ∆p/∆t = 1.95 Ns/0.020 s
10. m = 0.050 kg
F=?
F∆t = ∆p
F = 98 N
∆ v = +30 m/s
t = 0.10 s
F = ∆p/∆t
F∆t = m ∆ v = 0.050 kg (+30 m/s) = 1.5 kg m/s
F = 1.5 Ns/0.10 s
F = 15 N
4. c. change of momentum = impulse given to that object
∆p = F∆t F∆t = + 50 kg m/s = +50 N s
1. way
11. m = 0.41 kg
vi = 0
vf = 21 m/s t = 0.051 s
F=?
F∆t = m (v – u) = 0.41 kg (21 m/s - 0) = 8.61 kg m/s
F = 8.61 Ns/0.051 s
F = 170 N
p f = 12 kg m/s, upward
12. p = m v
5. Δp  p f  pi  mv  mu  m (v  u)  m Δv
it is easier if you draw final momentum first
p i = 12 kg m/s, downward
Δp  24 kg m/s, upward
b . p = (1210 kg)(51 m/s) = 62000 kg m/s
13. P1 = p2
2. way
choose positive direction. I chose upward to be positive.
Δp  p f - pi  12 kg m/s - (-12 kg m/s)  24 kg m/s
positive direction means upward, so
Δp  24 kg m/s, upward
3. way
do velocity first, then momentum
m1 v 1 = m 2 v 2
(1340 kg) v1 = (2680 kg)( 15 m/s, west)
v1 = 30 m/s, west
14. c. requires a smaller force
change of momentum = impulse given to that object (F∆t),
therefore the longer time interval has to be multiplied with the
smaller force in order to get the same change in momentum
15. m = 6.0 x 10–2 kg
∆p = ?
vi = 12 m/s, right
left (final direction) is positive
choose positive direction. I chose upward to be positive.

Δp  m (v  u)  6.0 x 10-2 kg
Δp  1.8 kg m/s, left
Δp  m (v  u)  1.012 - - 12  24 kg m/s
positive direction means upward, so
Δp  24 kg m/s, upward
vf = 18 m/s, left
or
 18m/s - - 12 m/s   1.8 kg m/s
Δp  - 1.8 kg m/s, right
16. 16. m = 0.30 kg
vi = 4.5 m/s downward vf = 4.2 m/s, upward
∆p = ?
p i = 1.35 kg m/s, downward
p f = 1.26 kg m/s, upward
6. e. Impulses (F∆t) acting on two bodies are equal, therefore change
in their momenta are equal.
7. impulse (F∆t) = momentum change (∆p)
choose positive direction. I chose upward to be positive.
∆p = pf – pi = 1.26 – (-1.35) = 2.6 kg m/s
∆p = 2.6 kg m/s, upward = - 2.6 kg m/s, downward
17. m = 0.2 kg
∆p = ?
vi = 40 m/s, right
vf = 60 m/s, left
left (final direction) is positive
Δp  pf - pi  4 kg m/s - (-6 kg m/s)  10 kg m/s
positive direction means upward, so
impulse  10 Ns, upward
8. momentum change (∆p) = impulse (F∆t)
F∆t = 1.0 Ns
so
∆p = 1.0 kg m/s
Δp  m (v  u)  0.2kg  40m/s - - 60 m/s   20 kg m/s
Δp  20 kg m/s, left
18. m = 0.15 kg
or
Δp  - 20 kg m/s, right
vi = 5.0 m/s, right
vf = 3 m/s, left
Δp  m (v  u)  0.15 kg  5m/s - - 3 m/s   1.2 kg m/s
Δp  1.2 kg m/s, left
or
Δp  - 1.2 kg m/s, right
2
25. c. The smaller fragment will have twice the speed
of the larger fragment.
Collisions, Ejections, Explosions
Law of conservation of momentum of the system
momentum of the system after collision is equal
to the momentum of the system before the collision
19.
3mu = 2mv2 + mv1
0 = 2v2 + v1
v1 = – 2v2
26.
mu = m1v1 + m2v2
0 = 500 v1 + 300
v1 = – 0.6 m/s
(m1 + m2) u = m1v1 + m2v2
negative sign means in the direction opposite to the swimmer
v1 = – 1 m/s
200 = 80 v1 + 120
v1 = 1 m/s, toward the shuttle
27.
20.
(m1 + m2) u = m1v1 + m2v2
mu = m1v1 + m2v2
0 = 70 v1 + 28
v1 = – 0.4 m/s
v1 = 3.7 m/s
9.15 = 0 + 2.50 v2
v1 = 3.7 m/s, toward the front of the ship
negative sign means in the direction opposite to the capsule
28.
21.
m1 u1 + m2 u2 = m1 v1 + m2 v2
3.0 - 10.0 = -3.0 + 5 v2
mu = m1v1 + m2v2
0 = 0.005 v1 + 1.1
v1 = - 0.8 m/s
v1 = 0.8 m/s to the left
v1 = – 220 m/s
29.
negative sign means in the direction opposite to the gun
22.
mu = m1v1 + m2v2
0 = 10.5 + 0.7 v2
m1 u1 + m2 u2 = (m1 + m2) v
20000 – 15000 = 2500 v
3.0 - 10.0 = -3.0 + 5 v2
v2 = – 15 m/s
v2 = – 15 m/s to the right
v = 2 m/s
v = 2 m/s, east
30.
23.
m1 u1 + m2 u2 = m1 v1 + m2 v2
7.0 u1 + 0 = 28 + 12
mu = m1v1 + m2v2
u1 = 5.7 m/s
u1 = 5.7 m/s to the right
210 = 0 + 30 v2
v2 = 7 m/s, right
The resulting change in speed of the cart is
7 m/s – 2 m/s = 5.0 m/s
24.
m1 u1 + m2 u2 = (m1 + m2 ) v
:
31.
0.35 u1 + 0 = 1.47
u1 = 4.2 m/s
u1 = 4.2 m/s to the north
m1 u1 + m2 u2 = m1 v1 + m2 v2
43.8 + 0 = 39.42 + 1.6 v2
v2 = + 2.7 m/s
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32. The only possible direction of momentum of the system of
two cars before the collision is direction 4 .
The law of conversation of momentum tells us that the change of
the momentum in the collision is zero, so the momentum of the
system must be the same after collision. As one car is heading east
the other car should head either north or in the direction 2 . Only
in those two cases resultant momentum of the system can be in
direction 4.
33.
toward the west. As the Earth rotates from west to east by itself
this additional momentum decreases the momentum of the Earth,
of course only slightly, very, very slightly (mass of all people,
animals, trains and trucks is very small compare to Earth mass)
38. b. the astronaut will never catch the first bounce.
Initially, momentum of the system astronaut + ball is zero. After
throwing the ball momentum of that system must be zero which
means that the momenta of the astronaut and ball are equal in
magnitude but opposite in direction. As the masses are equal, the
velocities are equal in magnitude but opposite in direction. The
ball will bounce off the wall with the same speed, and will never
be able to catch the astronaut.
39. F Δt = m Δv
Δv = F Δt/ m
Δv = (270 x 106)(800. N)(0.10 s)/(5.98 X1024 kg)
it is easier to use :
Δv = 3.6 X 10—15 m/s
Therefore, Earth’s motion would not be measurable.
40.
p1  p - p2
p = p1 + p2
F Δt = m Δv
components of the momentum of the 0.3 kg piece are:
0.8 kg m/s north and 1.0 kg m/s west
components of the velocity of the 0.3 kg piece are:
2.7 m/s north and 3.3 m/s west
34. in x direction:
m(3 m/s) + 0 = m v1 cos 300 + m (1.5 m/s) cos 600
v1 = 2.6 m/s
Δm
v = 5.25 x 103 m/s
= 136 kg/s
Δt
Δp Δ mv  Δm
F=
=
=
v
Δt
Δt
Δt
F = 136 kg/s   5.25 x 103 m/s 
F = 714 kN
Δ p1 + Δ p2 = 0
in x direction: obviously zero
2v = 10 sin 300 + 10 sin 300
41. No, the final momentum can equal zero only if the initial
momentum was zero. Because the cue ball was moving, its initial
momentum was not zero. Therefore, both balls cannot be at rest
after the collision.
Δ p = Δ (p1 + p2 ) = 0 , therefore
36. momentum before explosion = 0
momentum after explosion = 0
in y direction:
F = m Δv/Δt
F = (0.012 kg)(20.0 m/s)/( 2.0 X 10-4)
F = 1200 N
42. According to Newton's third law, the force of collision will be
the same on both the bug and the car, but will act in opposite
directions. The time during which the force acts is the same for
both, so the impulses on the bug and the car will be equal in
magnitude but opposite in direction. This means that they will
undergo equal and opposite changes in momentum. (It is
important to stress that their momenta are not the same, but that
they change by the same amount! The car loses as much
momentum as the bug gains in the collision.)
mathematically: momentum of the system doesn’t change what
means that change of the momentum of the system is zero:
3 = 0.866 v1 + 0.75
35.
Δt = Δd/v = = (0.0020 m)/(10.0 m/s) 2.0 X 10-4 s
v = 5 m/s
37. b. the earth would spin a bit slower.
Before they start to walk the momentum of the system Earth + all
people, animals, trains and trucks is zero. The law of conservation
of momentum tells us that the momentum of that system must be
zero when they start to walk.
Therefore momentum of the Earth must be equal, but opposite in
direction of the momentum of all people, animals, trains and
trucks.
As direction of momentum of all people, animals, trains and trucks
is towards the east the direction of momentum of the Earth is
or Δ p1 = – Δ p2
Because of the small mass of the bug, its acceleration will be very
large. Because of the large mass of the car, its acceleration will be
unnoticeable.
43. m = 50 kg
v = x/t = 400m/50s = 8 m/s
p = mv = 400 kg m/s
44. FΔt = mΔv
Δv = v – u = 0 – 2.0 = – 2.0 m/s
Δt = 0.5 s
F = mΔv /Δt
F = – 40 N in the direction opposite to initial velocity
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45.
Δp  m (v  u)  10 kg 
 4.0m/s-  -4.0 m/s  80 kg m/s
Δp  80 kg m/s, left
46. D
47. D
48.
F=
Δp
200
= slope =
= 20 N
Δt
10
49. A.
The change in its momentum is area under
the graph force vs. time of action
½ ( 50 x 0.275) + 50 x 0.175 + ½ ( 50 x 0.1) = 18.125 kg m/s
50. A. 0
no change in total momentum
Law of conservation of momentum in collisions:
If there is no net external force acting on the system, momentum
of the system doesn’t change.
51. B
The cart and the fan will remain at rest relative to the
surroundings, because the total momentum will remain zero.
There are only internal forces acting within the system. Don’t tell
me you don’t believe me.