A boy of $50 k g$ mass is running with a velocity of $2 m s^{- 1}$ . He jumped into a stationary cart of $2 k g$ while running. Find the velocity of the cart after the boy jumped into it.

$1.52 m s^{- 1}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.3 m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.92 m s^{- 1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3.25 m s^{- 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $1.92 m s^{- 1}$
Given:
Mass of the boy, $m_{1} = 50 k g$
Initial velocity of the boy, $u_{1} = 2 m s^{- 1}$
Mass of the cart, $m_{2} = 2 k g$
Initial velocity of the cart $u_{2} = 0$

Let the final velocity of cart be $v_{2}$
Since, boy jumped over cart, the final velocity $v_{1}$ of the boy will be equal to that of the cart.
Therefore, $v_{1} = v_{2}$

From law of conservation of momentum:
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$(50 \times 2) + (2 \times 0) = 50 v_{1} + 2 v_{2}$
$100 = 50 v_{2} + 2 v_{2} ∵ v_{1} = v_{2}$
$100 = 52 v_{2}$
$v_{2} = \frac{100}{52} = 1.92 m s^{- 1}$

Therefore, velocity of cart after jumping of boy over it is equal to $1.92 m s^{- 1}$ . Since, velocity has positive sign, the cart will go in the same direction of boy.

Suggest Corrections

Similar questions

A boy of 50 kg mass is running with a velocity of 2 m/s. He jumped into a stationary cart of 2 kg while running. Find the velocity of cart after the boy jumped into. (in m/s)

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 $\frac{m}{s}$ )