Question

Two persons standing on a floating boat run in succession along its length with a speed 4.2m/s relative to the boat and dive off from the end. The mass of each man is 80 kg and that of boat is 400 kg. lf the boat was initially at rest then (neglect friction)

• A. momentum of the boat with the second man is 300 kg-m/s.
• B. momentum of the boat with the second man is 336 kg-m/s.
• C. final velocity of the boat is 2.3 m/s.
• D. final velocity of the boat is zero.

Answer 1

Answer: (C)

final velocity of the boat is 2.3 m/s.

Let u be the velocity of the boat after the first man has jumped off. Then the velocity of the first man relative to the ground is (4.2 + u)m/s.
Since the momentum of the first man is equal and opposite to the momentum of boat and second man, we have
80(4.2 + u)$=$(400+80)u
$\Rightarrow u=0.84m/s$
After the first man has jumped off, the velocity of the boat with second man in it is 0.84 m/s
$\Rightarrow$ momentum of boat with the second man $=$(400+80)0.84
$=$403.2 kg-m/s
Let the velocity of the boat be v m/s after the second man has jumped off,
Momentum of boat $=$400v
Momentum of second man$=$80 $\times$ velocity of man relative to ground
$=$80(4.2 + v)
Momentum of boat and momentum of second man are in opposite direction
$\Rightarrow$ net momentum of boat and second man$=$400v-80(4.2+ v)
By conservation of momentum,
400v-80(4.2 + v)$=$403.2
$\Rightarrow v=2.30m/s$