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. $1.4v$ to the left
• C. $v$ to the left
• D. $0.4v$ to the right

Answer 1

Answer: (A)

$0.6v$ to the left

When A collides with B, the velocity of A becomes 0, and that of B becomes $v$.
When B moving with velocity $v$ towards C collides with C, since mass of C is greater, B will bounce back and start moving towards left with velocity. say $v_1$ and C will start moving with velocity, say $v_2$ towards left.

So, using law of conservation of momentum:
$mv=-m{v}_1+4m{v}_2$

$v_2=\frac{v+v_1}{4}$

Now using conservation of energy:

$\frac{1}{2}m{v}^2=\frac{1}{2}m{v_1}^2+\frac{1}{2}4m{v_2}^2$

Replacing $v_2$ in the equation we get:
$5{v_1}^2+2v{v}_1-3{v_2}^2=0$
Solving the quadratic equation we get:
$v_1=-0.6v$ i.e $0.6v$ towards left
When B collides with A again, A moves with velocity $v_1$, thus final velocity of block A is $0.6v$ towards left.