A truck moving on horizontal road east with velocity `20ms^(–1)` collides elastically with a light ball moving with velocity `25ms^(–1)` along west. The velocity of the ball just after collision

Options

A. 65 m`s^(–1)` towards east
B. 25 m`s^(–1)` towards wast
C. 65 m`s^(–1)` towards wast
D. 20 m`s^(–1)` towards east

Answer

To solve the problem of finding the velocity of the ball just after the elastic collision with the truck, we will use the principles of conservation of momentum and conservation of kinetic energy.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Velocity of the truck (V_t) = 20 m/s (East)
- Velocity of the ball (V_b) = -25 m/s (West, taken as negative for direction)
- Mass of the truck (M) = M (we can assume it to be M)
- Mass of the ball (m) = m (we can assume it to be m)

2. **Conservation of Momentum:**
The total momentum before the collision is equal to the total momentum after the collision.
\[ M \cdot V_t + m \cdot V_b = M \cdot V_{t}' + m \cdot V_{b}' \]
Substituting the values:
\[ M \cdot 20 + m \cdot (-25) = M \cdot V_{t}' + m \cdot V_{b}' \]

3. **Conservation of Kinetic Energy:**
Since the collision is elastic, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
\[ \frac{1}{2} M V_t^2 + \frac{1}{2} m V_b^2 = \frac{1}{2} M (V_{t}')^2 + \frac{1}{2} m (V_{b}')^2 \]
Substituting the values:
\[ \frac{1}{2} M (20)^2 + \frac{1}{2} m (-25)^2 = \frac{1}{2} M (V_{t}')^2 + \frac{1}{2} m (V_{b}')^2 \]

4. **Solving the Equations:**
From the momentum equation:
\[ 20M - 25m = MV_{t}' + mV_{b}' \]
From the kinetic energy equation:
\[ 200M + 312.5m = M(V_{t}')^2 + m(V_{b}')^2 \]

5. **Assuming the Masses:**
For simplicity, let’s assume M = 1 and m = 1 (the masses can be canceled out in the equations). This gives us:
\[ 20 - 25 = V_{t}' + V_{b}' \]
\[ -5 = V_{t}' + V_{b}' \]

6. **Using the Elastic Collision Formula:**
For elastic collisions, we can also use the formula for the final velocities:
\[ V_{b}' = \frac{(m_1 - m_2)}{(m_1 + m_2)} V_{1} + \frac{2m_2}{(m_1 + m_2)} V_{2} \]
Here, \( m_1 = M \) (truck) and \( m_2 = m \) (ball):
\[ V_{b}' = \frac{(1 - 1)}{(1 + 1)} \cdot 20 + \frac{2 \cdot 1}{(1 + 1)} \cdot (-25) \]
\[ V_{b}' = 0 + \frac{2}{2} \cdot (-25) = -25 \]

7. **Final Result:**
The velocity of the ball just after the collision is:
\[ V_{b}' = -25 \text{ m/s (West)} \]