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 with the wall, we will calculate the initial and final kinetic energies of the ball and then find the difference between them.

### Step 1: Calculate Initial Kinetic Energy
The initial kinetic energy (KE_initial) of the ball can be calculated using the formula:
\[ KE = \frac{1}{2} m v^2 \]
where:
- \( m = 1 \, \text{kg} \) (mass of the ball)
- \( v_1 = 10 \, \text{cm/s} = 10 \times 10^{-2} \, \text{m/s} = 0.1 \, \text{m/s} \) (initial velocity)

Substituting the values:
\[ KE_{\text{initial}} = \frac{1}{2} \times 1 \times (0.1)^2 = \frac{1}{2} \times 1 \times 0.01 = 0.005 \, \text{J} = 5 \times 10^{-3} \, \text{J} \]

### Step 2: Calculate Final Kinetic Energy
The final kinetic energy (KE_final) of the ball after it bounces back can be calculated similarly:
\[ KE_{\text{final}} = \frac{1}{2} m v'^2 \]
where:
- \( v' = 8 \, \text{cm/s} = 8 \times 10^{-2} \, \text{m/s} = 0.08 \, \text{m/s} \) (final velocity)

Substituting the values:
\[ KE_{\text{final}} = \frac{1}{2} \times 1 \times (0.08)^2 = \frac{1}{2} \times 1 \times 0.0064 = 0.0032 \, \text{J} = 3.2 \times 10^{-3} \, \text{J} \]

### Step 3: Calculate Loss in Energy
The loss in energy during the impact can be calculated by finding the difference between the initial and final kinetic energies:
\[ \text{Loss in Energy} = KE_{\text{initial}} - KE_{\text{final}} \]
Substituting the values:
\[ \text{Loss in Energy} = 5 \times 10^{-3} \, \text{J} - 3.2 \times 10^{-3} \, \text{J} = 1.8 \times 10^{-3} \, \text{J} \]

### Final Answer
The loss in energy of the ball during the impact is:
\[ \text{Loss in Energy} = 1.8 \times 10^{-3} \, \text{J} \]

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