If a body of mass m collides head on, elastically with velocity u with another identical boday at rest. After collision velocty of the second body will be

Options

A. zero
B. u
C. 2u
D. data insufficient.

Explanation

To solve the problem of two identical bodies colliding elastically, we will follow these steps:

### Step 1: Understand the Initial Conditions
We have two bodies of mass \( m \). Body A is moving with an initial velocity \( u \), and Body B is at rest. Thus, we can denote:
- Velocity of Body A before collision: \( V_A = u \)
- Velocity of Body B before collision: \( V_B = 0 \)

### Step 2: Apply Conservation of Momentum
According to the law of conservation of momentum, the total momentum before the collision must equal the total momentum after the collision.

The initial momentum \( P_{initial} \) is:
\[ P_{initial} = m \cdot u + m \cdot 0 = mu \]

Let \( V_A' \) be the velocity of Body A after the collision and \( V_B' \) be the velocity of Body B after the collision. The final momentum \( P_{final} \) is:
\[ P_{final} = m \cdot V_A' + m \cdot V_B' \]

Setting the initial momentum equal to the final momentum gives us:
\[ mu = m \cdot V_A' + m \cdot V_B' \]
Dividing through by \( m \):
\[ u = V_A' + V_B' \quad \text{(Equation 1)} \]

### Step 3: Apply Conservation of Energy
For an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

The initial kinetic energy \( KE_{initial} \) is:
\[ KE_{initial} = \frac{1}{2} m u^2 + 0 = \frac{1}{2} m u^2 \]

The final kinetic energy \( KE_{final} \) is:
\[ KE_{final} = \frac{1}{2} m (V_A')^2 + \frac{1}{2} m (V_B')^2 \]

Setting the initial kinetic energy equal to the final kinetic energy gives us:
\[ \frac{1}{2} m u^2 = \frac{1}{2} m (V_A')^2 + \frac{1}{2} m (V_B')^2 \]
Dividing through by \( \frac{1}{2} m \):
\[ u^2 = (V_A')^2 + (V_B')^2 \quad \text{(Equation 2)} \]

### Step 4: Solve the Equations
Now we have two equations:
1. \( u = V_A' + V_B' \)
2. \( u^2 = (V_A')^2 + (V_B')^2 \)

From Equation 1, we can express \( V_A' \) in terms of \( V_B' \):
\[ V_A' = u - V_B' \]

Substituting this into Equation 2:
\[ u^2 = (u - V_B')^2 + (V_B')^2 \]

Expanding the equation:
\[ u^2 = (u^2 - 2uV_B' + (V_B')^2) + (V_B')^2 \]
\[ u^2 = u^2 - 2uV_B' + 2(V_B')^2 \]

Cancelling \( u^2 \) from both sides:
\[ 0 = -2uV_B' + 2(V_B')^2 \]
Dividing through by 2:
\[ 0 = -uV_B' + (V_B')^2 \]
Factoring out \( V_B' \):
\[ V_B' (V_B' - u) = 0 \]

This gives us two solutions:
1. \( V_B' = 0 \) (which is not possible since Body B was initially at rest and must move after the collision)
2. \( V_B' = u \)

### Conclusion
Thus, the velocity of the second body (Body B) after the collision is:
\[ V_B' = u \]

### Final Answer
The velocity of the second body after the collision is \( u \).

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