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A bullet of mass m hits a block of mass ...

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

A

M' = M

B

M' = 2 M

C

M' lt lt M

D

M' gt gt M

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of maximum energy transfer when a bullet of mass \( m \) hits a block of mass \( M \), we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Problem**: - A bullet of mass \( m \) is moving with an initial velocity \( v \) and strikes a block of mass \( M \). We need to determine the condition under which the transfer of kinetic energy from the bullet to the block is maximized. 2. **Conservation of Momentum**: - According to the principle of conservation of momentum, the total momentum before the collision must equal the total momentum after the collision. - The initial momentum of the system is given by: \[ p_{\text{initial}} = mv \] - After the collision, let the bullet come to rest and the block move with velocity \( V \). The final momentum is: \[ p_{\text{final}} = MV \] - Setting the initial and final momentum equal gives: \[ mv = MV \] 3. **Kinetic Energy Before and After Collision**: - The initial kinetic energy (KE) of the bullet is: \[ KE_{\text{initial}} = \frac{1}{2} mv^2 \] - The final kinetic energy of the block (assuming the bullet transfers all its kinetic energy to the block) is: \[ KE_{\text{final}} = \frac{1}{2} MV^2 \] 4. **Condition for Maximum Energy Transfer**: - For maximum energy transfer, we want the bullet to transfer all its kinetic energy to the block. This occurs when the block moves with the same velocity as the bullet just before the collision. - Thus, we set \( V = v \) (the block moves with the same velocity as the bullet initially): \[ mv = Mv \implies M = m \] 5. **Conclusion**: - The transfer of energy is maximum when the mass of the block \( M \) is equal to the mass of the bullet \( m \). Therefore, the condition for maximum energy transfer is: \[ M = m \] ### Final Answer: The transfer of energy is maximum when the mass of the block \( M \) is equal to the mass of the bullet \( m \). ---
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