Question

A boy of mass $20\text{kg}$ on an initially stationary boat gets off the boat by jumping to the left in an exactly horizontal direction. Immediately after the jump, the boat of mass $40\text{kg}$ is observed to be moving to the right at speed $2\text{m/s}$. How much work did the boy do during the jump?

• A. $240\text{J}$
• B. $0\text{J}$
• C. $120\text{J}$
• D. $60\text{J}$

Answer 1

The correct option is A $240\text{J}$

Given, mass of the boy $\left(m\right)=20\text{kg}$
mass of the boat $\left(M\right)=40\text{kg}$
velocity of the boat $\left(v\right)=2\text{m/s}$
Let velocity of the boy w.r.t ground $=\left(v_1\right)$


There is no external force on the system (boy and boat), so linear momentum of the system remain constant in horizontal direction.

Initially, both the boy and boat are at rest. Applying conservation of linear momentum in horizontal direction.
$\therefore 0=Mv-m{v}_1$
$v_1=\frac{Mv}{m}$ ...(1)

Work done by the boy during the jump = Increase in total kinetic energy.
Since initially, both were at rest, final kinetic energy
$K.E.=$ $\frac{1}{2}m{v}_1^2+\frac{1}{2}M{v}^2$
$=\frac{1}{2}m{\left[\frac{Mv}{m}\right]}^2+\frac{1}{2}M{v}^2$
$=\frac{1}{2}\left[\frac{M^2{v}^2}{m}\right]+\frac{1}{2}M{v}^2$
$=\frac{1}{2}[\frac{M^2}{m}+M]v^2$
$=\frac{1}{2}[\frac{(40{)}^2}{20}+40]\times (2{)}^2$
$K.E=240\text{J}$
Therefore, work done by the boy during the jump is $240\text{J}$.