Question

A spring block system of mass 'm' and spring constant 'K' is placed at the friction less surface . A bullet of mass 'm' is fired towards the block of speed 'u' . If it is embedded in the block then the amplitude of the oscillation will be

• A. $\frac{(M+m)u}{m}\sqrt{\frac{m}{K}}$
• B. $\frac{mu}{(M+m)}\sqrt{\frac{m}{K}}$
• C. $\frac{mu}{(M+m)}\sqrt{\frac{M+m}{K}}$
• D. $\frac{(M+m)u}{m}\sqrt{\frac{k}{m}}$

Answer 1

Answer: (C)

$\frac{mu}{(M+m)}\sqrt{\frac{M+m}{K}}$

We know according to the law of conservation of momentum ,

$P_i=P_f$ //where $P_i$ is the initial momentum and $P_f$ is the final momentum .

$m$ is the mass of the bullet and $M$ is the mass of the Block. $u$ is the speed with which the bullet strikes and $K$ is the spring constant .

Let $V$ be the final velocity of the system after collision .

$mu=(M+m)V$

Or, $V=\frac{mu}{(M+m)}$

Now,

$\frac{1}{2}K{A}^2=\frac{1}{2}(M+m)V^2$ // where $A$ is the amplitude .

Or, $A^2=\frac{M+m}{K}\frac{m^2{u}^2}{(M+m{)}^2}$

Or ,$A^2=\frac{m^2{u}^2}{(M+m)K}$

Or, $A=\frac{mu}{M+m}\sqrt{\frac{M+m}{K}}$ .