The Physical World I

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 = 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.