Question

A man is standing on a cart which has a mass double that of the man as shown in the figure. Initially, the cart is at rest. Now, the man jumps horizontally with velocity $2\text{m/s}$ relative to the cart. Then work done by the man during the process of jumping will be

• A. $40\text{J}$
• B. $60\text{J}$
• C. $80\text{J}$
• D. $90\text{J}$

Answer 1

The correct option is C $80\text{J}$

Here, $m_1=120\text{kg},m_2=60\text{kg}$

Initially as the man and the cart are both at rest, $u_1=u_2=0\text{m/s}$.

Let us assume that the velocity of the cart is $v$ after the man jumps from the cart. It is given that the velocity of the man with respect to the cart is $2\text{m/s}.$
$v_{\text{man}}=v_{\left(\text{man, cart}\right)}-v_{\text{cart}}=2-v$

On applying the law of conservation of linear momentum before and after jumping we get,

$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$

$120\times 0+60\times 0=-120\times v+60\times (2-v)$

$0=-120v+120-60v$

$\Rightarrow 180v=120$

$\Rightarrow v=\frac{2}{3}\text{m/s}$

Now, speed of the man w.r.t ground

$=2-v=2-\frac{2}{3}=\frac{4}{3}\text{m/s}$

From the work energy theorem,

Work done by man = change in kinetic energy of system

$\Rightarrow W=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2-(\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2)$

$\Rightarrow W=\frac{1}{2}\times 120\times {\left(\frac{2}{3}\right)}^2+\frac{1}{2}\times 60\times {\left(\frac{4}{3}\right)}^2-(0+0)$

$\Rightarrow W=80\text{J}$