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A bullet of mass mm hits a stationary block of mass MM elastically. The transfer of energy is the maximum, when:

A

M=mM = m

B

M=2mM = 2m

C

M≪mM \ll m

D

M≫mM \gg m

Step-by-Step Solution

  1. Analyze Elastic Collision: In a head-on elastic collision between a projectile of mass m1m_1 (bullet, mm) and a stationary target of mass m2m_2 (block, MM), kinetic energy is conserved.
  2. Energy Transfer: The fractional kinetic energy transferred from the projectile to the target is given by: KblockKbullet=4m1m2(m1+m2)2=4mM(m+M)2\frac{K_{block}}{K_{bullet}} = \frac{4m_1 m_2}{(m_1 + m_2)^2} = \frac{4mM}{(m + M)^2}
  3. Maximize Transfer: This fraction is maximum (equal to 1 or 100%) when the numerator equals the denominator, i.e., 4mM=(m+M)24mM = (m + M)^2, which simplifies to (M−m)2=0(M - m)^2 = 0, implying M=mM = m.
  4. NCERT Reference: This corresponds to Case I in the text: 'If the two masses are equal... The first mass comes to rest and pushes off the second mass with its initial speed on collision.' Thus, the entire kinetic energy is transferred [Class 11 Physics, Ch 5, Sec 5.11.2, Case I].
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