A.) A 72 kg swimmer dives horizontally off a raft floating in a lake. The diver\

Question

A.) A 72 kg swimmer dives horizontally off a raft floating in a lake. The diver's speed immediately after leaving the raft is 3.8 m/s. If the time interval of the interaction between the diver and the raft is 0.25 s, what is the magnitude of the average horizontal force by diver on the raft?

A.  270 N

B.  68 N

C.  2300 N

D. 1100 N

B.) Suppose the mass of the raft in the previous problem is 500 kg. What is the raft's speed immediately after the 72 kg diver jumps off at 3.8 m/s?

A. 0.14 m/s

B. 0.27 m/s

C. 0.55 m/s

D. 3.8 m/s

Explanation / Answer

A. Change in momentum = m x v - 0 = mv = 72 x 3.8 = 273.6 kgm/s

Change in momentum = force x time

force = Change in momentum/time = 273.6/0.25 = 1100 N. Hence option D is correct.

B. By Newton's third law momentum given to boat = momentum of diver = 72 x 3.8 = 273.6 Kgm/s

So velocity of boat = 273.6/mass = 273.6/500 = 0.55 m/s

Hence option C is correct.

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