Question

Three objects \(\text{A, B}\) and \(\text C\) are kept in a straight line on a frictionless horizontal surface. The masses of \(\text{A, B}\) and \(\text C\) are \(m,~2~m\) and \(2~m\) respectively. \(\text A\) moves towards \(\text B\) with a speed of \(9~\text{m/s}\) and makes an elastic collision with it. Thereafter \(\text B\) makes a completely inelastic collision with \(\text C\). All motions occur along same straight line. The final speed of \(\text C\) is :

• A. $4\text{m/s}$
• B. $3\text{m/s}$
• C. $6\text{m/s}$
• D. $9\text{m/s}$

Answer 1

The correct option is B $3\text{m/s}$

Collision between $\text{A}$ and $\text{B}$ is elastic, so both momentum and energy will be conserved.

Applying momentum conservation for above collision,

$m\times 9=m{V}_1+\left(2m\right)V_2$

$\Rightarrow V_1+2V_2=9......\left(1\right)$

Now, since the collision between $m$ and $2m$ is elastic, so

coefficient of restitution, $e=1=\frac{V_2-V_1}{9}$

$\Rightarrow -V_1+V_2=9.........\left(2\right)$

From eqs. $\left(1\right)\&\left(2\right)$, we get

$V_2=6\text{m/s},V_1=-3\text{m/s}$

Collision between $\text{B}$ and $\text{C}$ is inelastic, so after the collision both the blocks will stick together and move with velocity $V$.

From momentum conservation,

$2m\times 6=4mV$

$\therefore V=3\text{m/s}$

Therefore, the correct answer is option $\left(\text{D}\right)$.