A bullet of mass m hits a block of mass M. The transfer of energy is maximum when

Options

A. `m gt gt M `
B. `M gt gt m`
C. `M =2m`
D. `M=m`

Answer

To solve the problem of determining when the transfer of energy is maximum when a bullet of mass \( m \) hits a block of mass \( M \), we can follow these steps:

### Step 1: Understand the System
We have a bullet of mass \( m \) moving with an initial velocity \( u \) that collides with a block of mass \( M \) which is initially at rest. After the collision, the bullet embeds itself into the block.

### Step 2: Conservation of Momentum
Using the principle of conservation of momentum, we can express the momentum before and after the collision. The initial momentum \( p_i \) is given by:
\[ p_i = mu + 0 = mu \]
The final momentum \( p_f \) after the collision, when the bullet and block move together with a velocity \( v \), is:
\[ p_f = (M + m)v \]
Setting these equal gives us:
\[ mu = (M + m)v \]
From this, we can solve for \( v \):
\[ v = \frac{mu}{M + m} \]

### Step 3: Calculate the Change in Kinetic Energy
The change in kinetic energy (\( \Delta KE \)) can be calculated using the formula:
\[ \Delta KE = KE_{final} - KE_{initial} \]
The initial kinetic energy \( KE_{initial} \) is:
\[ KE_{initial} = \frac{1}{2}mu^2 \]
The final kinetic energy \( KE_{final} \) after the collision is:
\[ KE_{final} = \frac{1}{2}(M + m)v^2 \]
Substituting \( v \) from Step 2 into the final kinetic energy:
\[ KE_{final} = \frac{1}{2}(M + m)\left(\frac{mu}{M + m}\right)^2 \]
Simplifying this gives:
\[ KE_{final} = \frac{1}{2}(M + m)\frac{m^2u^2}{(M + m)^2} = \frac{1}{2}\frac{m^2u^2}{M + m} \]

### Step 4: Find the Change in Kinetic Energy
Now, substituting back into the change in kinetic energy:
\[ \Delta KE = \frac{1}{2}\frac{m^2u^2}{M + m} - \frac{1}{2}mu^2 \]
Factoring out \(\frac{1}{2}u^2\):
\[ \Delta KE = \frac{1}{2}u^2\left(\frac{m^2}{M + m} - m\right) \]
This simplifies to:
\[ \Delta KE = \frac{1}{2}u^2\left(\frac{m^2 - m(M + m)}{M + m}\right) = \frac{1}{2}u^2\left(\frac{m^2 - mM - m^2}{M + m}\right) = \frac{1}{2}u^2\left(\frac{-mM}{M + m}\right) \]

### Step 5: Determine Maximum Energy Transfer
The transfer of energy is maximum when the expression for \(\Delta KE\) is maximized. This occurs when the ratio of the masses \( \frac{m}{M} \) is equal to 1, meaning \( M = m \).

### Conclusion
Thus, the transfer of energy is maximum when the mass of the block \( M \) is equal to the mass of the bullet \( m \).

### Final Answer
The transfer of energy is maximum when \( M = m \).

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