A man of 50 kg is standing at one end on a boat of length 25 m and mass 200 kg . If he starts running and reaches the other end, he has a speed 2 m s- 1 with respect to the boat. The final speed of the boat is ( in .m s-1)

Question

A man of 50 kg is standing at one end on a boat of length 25 m and mass 200 kg . If he starts running and reaches the other end, he has a speed 2 m s- 1 with respect to the boat. The final speed of the boat is ( in .m s-1) (A) (2/5) (B) (2/3) (C) (8/5) (D) (8/3)

Tardigrade Admin · Accepted Answer

Let v be the speed of the boat with respect to the water, then from conservation of linear momentum. -(200 + 50)v+50× 2=50× 0+200× 0 250v=-100 v=-(100/250)=-(2/5) m s- 1

Q. A man of $50 k g$ is standing at one end on a boat of length $25 m$ and mass $200 k g$ . If he starts running and reaches the other end, he has a speed $2 m s^{- 1}$ with respect to the boat. The final speed of the boat is ( in $m s^{- 1})$

$\frac{2}{5}$

$\frac{2}{3}$

$\frac{8}{5}$

$\frac{8}{3}$

Let $v$ be the speed of the boat with respect to the water, then from conservation of linear momentum.
$- (200 + 50) v + 50 \times 2 = 50 \times 0 + 200 \times 0$
$250 v = - 100$
$v = - \frac{100}{250} = - \frac{2}{5} m s^{- 1}$

Q. A man of 50kg is standing at one end on a boat of length 25m and mass 200kg . If he starts running and reaches the other end, he has a speed 2ms−1 with respect to the boat. The final speed of the boat is ( in ms−1)

52​

32​

58​

38​