A shot of mass m penetrates a thickness t of a fixed plate of mass M.If M were free to move and the resistance supposed to be uniform, find the thickness penetrated.

Options

A. `( Mt)/( M+m)`
B. `(mt)/(M+m)`
C. `(t)/(M+m)`
D. `(t)/( M+2m)`

Answer

To solve the problem of how much thickness a shot of mass \( m \) penetrates into a plate of mass \( M \) when \( M \) is free to move, we can apply the principles of conservation of momentum and the work-energy theorem. Here is a step-by-step solution:

### Step 1: Understand the scenario
Initially, we have a shot of mass \( m \) moving with a velocity \( v \) towards a plate of mass \( M \). In the first case, the plate is fixed, and the shot penetrates a thickness \( t \). In the second case, the plate is free to move.

### Step 2: Apply the work-energy theorem for the first case
For the first case where the plate is fixed, the work done by the friction force (which is negative) equals the change in kinetic energy of the shot.

The work done by the friction force can be expressed as:
\[ W = -F_f \cdot t \]
where \( F_f \) is the friction force and \( t \) is the thickness penetrated.

The initial kinetic energy of the shot is:
\[ KE_i = \frac{1}{2} mv^2 \]
and the final kinetic energy when it stops is:
\[ KE_f = 0 \]

By the work-energy theorem:
\[ -F_f \cdot t = KE_f - KE_i \]
\[ -F_f \cdot t = 0 - \frac{1}{2} mv^2 \]
\[ F_f \cdot t = \frac{1}{2} mv^2 \]

### Step 3: Apply the conservation of momentum for the second case
In the second case, both masses \( m \) and \( M \) are moving together after the collision. Let \( v' \) be their common velocity after the collision. By conservation of momentum:
\[ mv = (M + m)v' \]
From this, we can solve for \( v' \):
\[ v' = \frac{mv}{M + m} \]

### Step 4: Apply the work-energy theorem for the second case
Now, we apply the work-energy theorem again. The work done by the friction force when the shot penetrates a thickness \( t' \) is:
\[ -F_f \cdot t' = KE_f - KE_i \]
The initial kinetic energy is:
\[ KE_i = \frac{1}{2} mv^2 \]
The final kinetic energy when the shot and the plate are moving together is:
\[ KE_f = \frac{1}{2}(M + m)v'^2 = \frac{1}{2}(M + m)\left(\frac{mv}{M + m}\right)^2 = \frac{m^2v^2}{2(M + m)} \]

Setting up the equation:
\[ -F_f \cdot t' = \frac{m^2v^2}{2(M + m)} - \frac{1}{2} mv^2 \]
Substituting \( F_f = \frac{mv^2}{2t} \) from the first case:
\[ -\frac{mv^2}{2t} \cdot t' = \frac{m^2v^2}{2(M + m)} - \frac{1}{2} mv^2 \]

### Step 5: Solve for \( t' \)
Rearranging gives:
\[ t' = \frac{m}{M + m} \cdot t \]

### Final Answer
Thus, the thickness penetrated when the plate is free to move is:
\[ t' = \frac{mt}{M + m} \]