Spring 2005

MWF 11:30

**Instructor**: Dr. Bruce Law, CW
327, Tel: 532-1618.

# REVIEW QUESTIONS OF Home Work 6

# R7. The longer cannon because the force acts
over a longer time. (Recall that Ft = change in momentum.)

# R10. It is good to have your hands extended
because then you can stop the ball over a longer period of time – it then
requires less force to stop the ball. (Again you are using Ft = change in
momentum.)

# R18. To conserve in physics means to keep
constant.

# R21. In an elastic collision, the objects
bounce off each other. In an inelastic collision, the objects stick together.
Momentum is conserved in both types of collisions.

# R22. Car B has the same initial speed as car
A in order to conserve momentum.

# R23. Momentum before = momentum after.

# Therefore Mv(before) = 2Mv(after).

# Hence v(after) =
v(before)/2 i.e. speed after the
collision is half the initial speed of car A.

#

# EXERCISES OF Home Work 6

## E1. Supertankers are very massive, therefore even
at a modest speed they have a very large momentum. In order to change this
momentum one must apply a modest force (i.e. friction from water) over a very
long time (i.e. 25km from port).

## E4. Change in momentum = force x time. The change in momentum to stop the
gymnast remains the same. The best way to stop the gymnast is therefore to
apply a small force over a long time by using thick floor mats.

**E8. On the
moon the lunar vehicle's mass m remains the same, hence it's momentum p=mv remains the same. **

**E16. Change in
momentum = force x time. The change in momentum to stop the punch remains the
same. If you apply the punch with a bare fist, the interaction time is short
and therefore the force is large. However, with a boxing glove the interaction
time is long and therefore the force is smaller. **

** **

# PROBLEMS OF Home Work 6

**P2. Change in
momentum = mv(final) - mv(initial) = Ft.
Therefore the braking force F=(mv(final)-mv(initial))/t = (0-1000x20)/10 = -2000N. **

**P4. Question
(a) can be answered. Impulse = Ft = change in momentum. For this situation the
change in momentum = (0-1000x30) = -30,000kg m/s. Question (b) cannot be
answered because the interaction time t with the ground is not known. **

**P5. Momentum
of the caught ball is (0.15kg)(40m/s) = 6.0 kg m/s.**

**(a)
****The
impulse to produce this change of momentum has the same magnitude, i.e. 6.0 kg
m/s.**

**(b)
****From Ft = ****D****mv****, F = ****D****mv****/t = (6.0 kg m/s)/0.03s = 200N.**

**P9. Momentum(before) = momentum(after)**

**(5kg)(1m/s) +
(1kg)V = 0**

**Therefore V =
-5m/s.**

**So if the
little fish approaches the big fish at 5m/s, the momentum after lunch will be
zero.**

**P10. The
initial total momentum for this problem is p(i) = 0 kg m/s. The final total total
momentum p(f) = m(super)v(super)+1000m(super)v(asteroid) where we have used the
fact that m(asteroid) = 1000m(super). According to conservation of momentum p(i) = p(f), therefore, after a
little algebra Superman's speed would be v(super) = -1000v(asteroid) =
-1000x800 = -800,000 m/s. **