A boy is running with a velocity of $2 m s^{- 1}$ . He jumps over a stationary cart of $2 k g$ while running. If the mass of the boy is of $20 k g$ , find the velocity at which the cart moves after he jumps over the cart. (Neglect friction between the cart and ground)

1.81 m s^{- 1}

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1.92 m s^{- 1}

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2.44 m s^{- 1}

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1.3 m s^{- 1}

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Solution

The correct option is A $1.81 m s^{- 1}$
Given mass of the boy, $m_{1} = 20 k g$
Initial velocity of boy, $u_{1} = 2 m s^{- 1}$
Given mass of the cart, $m_{2} = 2 k g$
Initial velocity of the cart, $u_{2} = 0$ (Initially at rest)
Let $v_{1}$ be the final velocity of the boy and $v_{2}$ be the final velocity of the cart.

Since,the boy jumped over the cart, the final velocity $v_{1}$ of boy will be equal to that of the cart.
Therefore $v_{1} = v_{2}$
According to law of conservation of momentum:
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
Sustituting the values , we get,
$20 \times 2 + 2 \times 0 = 20 \times v_{1} + 2 \times v_{2}$
$\Rightarrow$ $40 = 20 \times v_{2} + 2 \times v_{2}$
$\Rightarrow$ $40 = (20 + 2) \times v_{2}$
$\Rightarrow$ $\frac{40}{22} = v_{2}$
$\Rightarrow$ $v_{2} = 1.81 m s^{- 1}$
Therefore, velocity of the cart after he jumps over it is equal to $1.81 m s^{- 1}$ .

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