A ball of mass m = 1 kg rolling without slipping, strikes a smooth vertical wall and comes back. Before the impact the velocity of the ball is `v_(1) = 10` cm/s and just after the impact `v_(1) = 8` cm/s. Find the loss in energy of ball during the impact.

Answer

To find the loss in energy of the ball during the impact, we will calculate the change in kinetic energy before and after the impact.

### Step-by-step Solution:

1. **Identify the mass and velocities**:
- Mass of the ball, \( m = 1 \, \text{kg} \)
- Initial velocity before impact, \( v_1 = 10 \, \text{cm/s} = 0.1 \, \text{m/s} \) (conversion from cm/s to m/s)
- Final velocity after impact, \( v_2 = 8 \, \text{cm/s} = 0.08 \, \text{m/s} \) (conversion from cm/s to m/s)

2. **Calculate the initial kinetic energy (KE_initial)**:
\[ KE_{\text{initial}} = \frac{1}{2} m v_1^2 \]
Plugging in the values:
\[ KE_{\text{initial}} = \frac{1}{2} \times 1 \times (0.1)^2 = \frac{1}{2} \times 1 \times 0.01 = 0.005 \, \text{J} \]

3. **Calculate the final kinetic energy (KE_final)**:
\[ KE_{\text{final}} = \frac{1}{2} m v_2^2 \]
Plugging in the values:
\[ KE_{\text{final}} = \frac{1}{2} \times 1 \times (0.08)^2 = \frac{1}{2} \times 1 \times 0.0064 = 0.0032 \, \text{J} \]

4. **Calculate the loss in kinetic energy (ΔKE)**:
\[ \Delta KE = KE_{\text{initial}} - KE_{\text{final}} \]
Plugging in the values:
\[ \Delta KE = 0.005 \, \text{J} - 0.0032 \, \text{J} = 0.0018 \, \text{J} \]

5. **Conclusion**:
The loss in energy of the ball during the impact is \( 0.0018 \, \text{J} \) or \( 1.8 \times 10^{-3} \, \text{J} \).