Question

Three blocks are initially placed as shown in the figure. Block $A$ has mass $m$ and initial velocity $v$ to the right. Block $B$ with mass $m$ and block $C$ with mass $4m$ are both initially at rest. Neglect friction. All collisions are elastic. The final velocity of block $A$ is

• A. $0.6v$ to the left
• B. $0.4v$ to the left
• C. zero
• D. $0.4v$ to the right

Answer 1

The correct option is A $0.6v$ to the left


Since the collision between $A$ and $B$ is elastic, velocity of $B$ after collision,
$V_B=v$ {Since, $m_A=m_B=m$ }

Now, for collision between $B$ and $C$ :
Let final velocity of B be $v_B$ (right)
final velocity of C be $v_C$ (right)

Applying linear momentum conservation between $B$ and $C$:
$m_B{u}_B+m_C{u}_C=m_B{v}_B+m_C{v}_C$
$\Rightarrow mv=m{v}_B+4m{v}_C$
$v=v_B+4v_C$ ...(i)

Also, $e=1=\frac{v_{separation}}{v_{approach}}=\frac{v_C-v_B}{v-0}$
$\Rightarrow v_C-v_B=v$ ...(ii)

Now from (i) and (ii), $v_C=\frac{2v}{5}$ (towards right)
& $v_B=-\frac{3v}{5}$ (towards left)

Again, B collides with stationary 'A' and exchanges velocity.
$v_{A_{final}}=-\frac{3v}{5}$ (towards left)