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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A boy is running with a velocity of 2 ms−1. He jumps over a stationary cart of 2 kg while running. If the mass of the boy is of 20 kg, find the velocity at which the cart moves after he jumps over the cart. (Neglect friction between the cart and ground)

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)