Question

A boy of $50$ kg mass is running with a velocity of $2$ $\frac{m}{s}$. He jumped into a stationary cart of $2$ kg while running. Find the velocity of the cart after the boy jumped into it. (in $\frac{m}{s}$)

• A. $1.52$ $\frac{m}{s}$
• B. $2.3$ $\frac{m}{s}$
• C. $1.92$ $\frac{m}{s}$
• D. $3.25$ $\frac{m}{s}$

Answer 1

The correct option is C

$1.92$ $\frac{m}{s}$

Given,

Mass ($m_1$) of boy $=50$ kg

Initial Velocity ($u_1$) of boy $=2$ $\frac{m}{s}$

Mass ($m_2$) of cart $=2$ kg

Initial Velocity ($u_2$) of cart $=0$

We have to calculate final velocity of the cart, $v_2$.

Since, boy jumped into the cart, the final velocity ($v_1$) of the boy will be equal to that of the cart.

Therefore, $v_1=v_2$

Conservation of momentum:

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

$\Rightarrow 50\times 2+2\times 0=50\times v_1+2\times v_2$

Since, $v_1=v_2$

$\Rightarrow 100=50\times v_2+2\times v_2$

$\Rightarrow 100=52v_2$

$\Rightarrow v_2=\frac{100}{52}=1.92$ $\frac{m}{s}$

Therefore, velocity of the cart after the boy jumped into it is equal to $1.92$ $\frac{m}{s}$. Since, velocity has positive sign, thus, cart will go in the same direction as that of boy.