A `60 kg` boy lying on a surface of negliguble friction throws horizontally a stone of mass `1 kg` with a speed of `12 m//s` away from him. As a result with what kinetic energy he moves back ?

Options

A. 2.4 J
B. 72 J
C. 1.2 J
D. 36 J

Explanation

To solve the problem step by step, we will use the principle of conservation of momentum and the formula for kinetic energy.

### Step 1: Understand the system
The boy has a mass of 60 kg and throws a stone of mass 1 kg with a speed of 12 m/s. Initially, both the boy and the stone are at rest, so their initial momentum is zero.

### Step 2: Apply the conservation of momentum
According to the law of conservation of momentum, the total momentum before throwing the stone must equal the total momentum after throwing the stone.

- Initial momentum (before throwing) = 0
- Final momentum (after throwing) = momentum of the boy + momentum of the stone

Let \( V \) be the velocity of the boy after throwing the stone. The momentum of the stone after being thrown is:
\[ \text{Momentum of stone} = \text{mass of stone} \times \text{velocity of stone} = 1 \, \text{kg} \times 12 \, \text{m/s} = 12 \, \text{kg m/s} \]

The momentum of the boy moving in the opposite direction is:
\[ \text{Momentum of boy} = -\text{mass of boy} \times V = -60 \, \text{kg} \times V \]

Setting the total final momentum equal to the initial momentum:
\[ 0 = -60V + 12 \]

### Step 3: Solve for V
Rearranging the equation gives:
\[ 60V = 12 \]
\[ V = \frac{12}{60} = 0.2 \, \text{m/s} \]

### Step 4: Calculate the kinetic energy of the boy
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} m V^2 \]
where \( m \) is the mass and \( V \) is the velocity.

Substituting the values for the boy:
\[ KE = \frac{1}{2} \times 60 \, \text{kg} \times (0.2 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \times 60 \times 0.04 \]
\[ KE = 30 \times 0.04 \]
\[ KE = 1.2 \, \text{J} \]

### Final Answer
The kinetic energy with which the boy moves back is **1.2 Joules**.

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