Two wooden plank of mass `M_(1)= 1 kg, M_(2) = 2.98 kg` smooth surface. A bullet of mass `m = 20 gm` strikes the block `M_(1)` and pierces through it, then strikes the plank `B` and sticks to its. Consequently both the planks move with equal velocities. Find the percentage change in speed of the bullet when it escapes from the first block.

Explanation

To solve the problem, we will use the principle of conservation of momentum. Let's break down the solution step by step.

### Step 1: Understand the system and initial conditions
We have two wooden planks, \( M_1 = 1 \, \text{kg} \) and \( M_2 = 2.98 \, \text{kg} \), and a bullet of mass \( m = 20 \, \text{g} = 0.02 \, \text{kg} \). The bullet strikes plank \( M_1 \) and then pierces through it, subsequently hitting plank \( M_2 \) and sticking to it.

### Step 2: Apply conservation of momentum for the bullet and plank \( M_1 \)
When the bullet strikes plank \( M_1 \), we can write the conservation of momentum as:
\[ m u = M_1 v + m u_1 \quad \text{(1)} \]
where:
- \( u \) = initial velocity of the bullet,
- \( v \) = velocity of plank \( M_1 \) after the bullet passes through,
- \( u_1 \) = velocity of the bullet after passing through plank \( M_1 \).

### Step 3: Apply conservation of momentum for the bullet and plank \( M_2 \)
After the bullet exits plank \( M_1 \), it strikes plank \( M_2 \) and sticks to it. The conservation of momentum for this scenario is:
\[ m u_1 = (M_2 + m) V \quad \text{(2)} \]
where \( V \) is the common velocity of both planks after the bullet sticks to plank \( M_2 \).

### Step 4: Relate the velocities
From equation (2), we can express \( V \):
\[ V = \frac{m u_1}{M_2 + m} \]

### Step 5: Substitute \( V \) into equation (1)
Substituting \( V \) into equation (1):
\[ m u = M_1 v + m u_1 \]
We can express \( v \) in terms of \( u \) and \( u_1 \):
\[ v = \frac{m u - m u_1}{M_1} \]

### Step 6: Find the relationship between \( u \) and \( u_1 \)
We can express \( u_1 \) in terms of \( u \):
\[ u_1 = \frac{M_1 u + m u - M_1 V}{m} \]
Substituting \( V \) from step 4 into this equation gives us a relationship between \( u \) and \( u_1 \).

### Step 7: Calculate the percentage change in speed
The percentage change in speed of the bullet is given by:
\[ \text{Percentage Change} = \frac{u_1 - u}{u} \times 100 \]

### Step 8: Substitute values and simplify
Substituting the values:
- \( M_1 = 1 \, \text{kg} \)
- \( M_2 = 2.98 \, \text{kg} \)
- \( m = 0.02 \, \text{kg} \)

We find:
\[ \text{Percentage Change} = \frac{-M_1}{M_2 + m} \times 100 \]
Calculating this gives:
\[ \text{Percentage Change} = \frac{-1}{2.98 + 0.02} \times 100 = \frac{-1}{3} \times 100 \approx -33.33\% \]

Thus, the percentage change in speed of the bullet when it escapes from the first block is approximately **33.33%**.