JEE Main · PhysicsHard

$A block of mass M is connected to a spring of constant k on a smooth horizontal surface. A bullet of mass m moving with velocity v hits the block and gets embedded. What is the amplitude of the resulting oscillation?$

Show correct answer & step-by-step solution

Correct answer: A — $\frac{m v}{k ( M + m )}$

Solution

$By conservation of momentum, (M + m) V = m v, so V = \frac{m v}{M + m} .$
$The total energy of the system is E = \frac{1}{2} (M + m) V^{2} = \frac{1}{2} k A^{2} .$
$Substituting V, we get \frac{1}{2} (M + m) (\frac{m v}{M + m})^{2} = \frac{1}{2} k A^{2} .$
$Solving for A, we get A = \frac{m v}{k ( M + m )} .$
$Hence the answer is (A).$