Question

To get off a frozen, frictionless lake, a 70.0 kg person takesoff a 0.150 kg shoe and throws it horizontally, directly away fromthe shore with a speed of 2.00 m/s.

If the person is 5.00 m from the shore, how long does it takefor him to reach it?

Answer 1

You have to apply the law of conservation of momentum to tackel this problem. The man on the frozen ice is at rest before throwing the shoe. Thus the momentum of the system is zero before throwing the shoe. From the law of conservation of linear momentum the total momentum before throwing the shoe is equal to the total momentum after throwing the shoe.
Momentum of the shoe = momentum of the man on ice
mv = MV; V = mv/M = 0.5*2/70 = 1/70 m/s.
Time required to travel the distance S = 5m is given by S = Vt; Thus t = S/V = 5/(1/70) = 5*70 = 350s = 5 min 50 s.