# The Physical World I

Spring 2005
MWF 11:30

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

# EXERCISES OF Home Work 6

## 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 = Dmv, F = Dmv/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.