Mid-term Review : Momentum
Problems: 2, 16, 19a, 20, 21, 22, 23
2.
Before the collision:
Pa = MaVa = (2E3 kg) * (3m/s) = 6E3 kg m/sec (Moving to the right)
Pb = -6E3 kgm/sec = MbVb = Mb(-1.5m/sec) (Moving to the left)
After the collision:
Pa + Pb = 0
a) Pa = MaVa = (2E3 kg) * (3m/s) = 6E3 kg m/sec (Moving to the right). Technically you do not
need to indicate the direction (since it says give the magnitude).
b) Draw a 6cm line pointing to the right.
c) The momentum of the cart after the collision is zero, since the momentum of each cart is of
equal magnitude but they are moving in different directions.
Problems 15-17. Do only Part 16.
15. Cart 1: m = 1000kg, v = 6m/s
Cart 2: m = 5000kg, v = 0m/sec.
P(cart1, initial) = 6000kgm/sec.
P(cart2, initial) = 0.
P(cart1+cart2) = m(cart1)(v(cart1,initial) + m(cart2)v(cart2, initial) = (m(cart1)+m(cart2))v final
(1000kg)(6m/sec) + 0 = 6000kgm/sec = (1000kg + 5000kg)v final = (6000kg)v final
16. Vfinal = (6000kgm/sec)/6000kg = 1m/sec
Problems 18-19. Do only 19a.
19 a. P(initial) = P(child) + P(sled)
= m(child)v(child) + p(sled)v(sled) = 50kg * 6m/sec + 10kg * 0 m/sec = 300kgm/sec
P(final) = P(initial) = (50kg + 10kg) * v(final)
300 kgm/sec = (60kg) * v(final)  (300 kgm/sec)/(60kg) = 5m/sec
20. P(initial) = mv = 1000kg * 15m/sec = 15000kgm/sec.
J = 6000Nsec =  p. THIS IS THE ANSWER.
P(final) = 15000kgm/sec + 6000Nsec = 15000kgm/sec + 6000kgm/sec = 21000kgm/sec, so the
change in velocity could be calculated as
21000kgm/sec / (1000kg) = 21m/sec. Be careful to provide the answer the question is asking for.
21. This kind of problem could be on the mid-term. Draw a best-fit line USING A STRAIGHT EDGE. Do
not just connect the dots.
The equation T^2 = (4 Pi^2 / g) *l is in the form y = mx + b, where y = T^2, l = x, and (4 Pi^2 / g) is the
slope. b = 0 in this case.
22 and 23.
22. Ball is fired to the east, cannon recoils to the west.
(For every action, there is an equal and opposite reaction.
Initial momentum is zero. Final momentum is zero. If the cannonball ends up with a
momentum to the east, the cannon must acquire a momentum to the west.)
23. P(initial) = 0 = P(final).
P(ball, initial) + P(cannon, initial) = P(ball, final)+P(cannon, final) = 0
P(ball, final) + P(cannon, final) = m(ball)v(ball,final) + m(cannon)v(cannon, final)
P(ball, final) + P(cannon, final) = 2.4E4kgm/sec + 1E3kg * v(cannon, final) = 0
 2.4E3 kgm/sec = -1E3kg * v(cannon, final)
 2.4E3kgm/sec /1E3kg = -v(cannon, final)
 -2.4 m/sec = v(cannon, final)