Two identical blocks A and B of mass M each are kept on each other on a smooth horizontal plane. There exists friction between A and B. A bullet of mass m hits the the lower block with a horizontal velocity v and goes embedded into it. Find the work done by friction between A and B.

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the system
We have two identical blocks A and B, each with mass M, stacked on top of each other on a smooth horizontal plane. A bullet of mass m strikes the lower block B with a horizontal velocity v and embeds itself into it.

### Step 2: Apply the principle of conservation of momentum
Before the collision, the momentum of the system is given by the bullet's momentum since the blocks are at rest. After the bullet embeds into block B, we can consider the combined mass of block B and the bullet.

The initial momentum (before the collision) is:
\[ P_{\text{initial}} = mv \]

After the bullet embeds into block B, the total mass moving together is \( M + m \) (mass of block B plus the mass of the bullet). Let the common velocity after the collision be \( v_0 \).

Using conservation of momentum:
\[ mv = (M + m)v_0 \]

From this, we can solve for \( v_0 \):
\[ v_0 = \frac{mv}{M + m} \]

### Step 3: Calculate the initial and final kinetic energy
The initial kinetic energy (KE_initial) of the system is only due to the bullet:
\[ KE_{\text{initial}} = \frac{1}{2} mv^2 \]

The final kinetic energy (KE_final) of the system after the collision (both blocks and the bullet moving together) is:
\[ KE_{\text{final}} = \frac{1}{2}(M + m)v_0^2 \]

Substituting \( v_0 \):
\[ KE_{\text{final}} = \frac{1}{2}(M + m)\left(\frac{mv}{M + m}\right)^2 \]
\[ KE_{\text{final}} = \frac{1}{2}(M + m)\frac{m^2v^2}{(M + m)^2} \]
\[ KE_{\text{final}} = \frac{1}{2}\frac{m^2v^2}{M + m} \]

### Step 4: Calculate the work done by friction
The work done by friction (W_friction) between blocks A and B can be found using the change in kinetic energy of block A. Since block A starts from rest, its initial kinetic energy is 0.

Thus, the work done by friction is:
\[ W_{\text{friction}} = KE_{\text{final}} - KE_{\text{initial}} \]
\[ W_{\text{friction}} = \frac{1}{2}\frac{m^2v^2}{M + m} - 0 \]
\[ W_{\text{friction}} = \frac{1}{2}\frac{m^2v^2}{M + m} \]

### Final Answer
The work done by friction between blocks A and B is:
\[ W_{\text{friction}} = \frac{1}{2}\frac{m^2v^2}{M + m} \]

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