MCQ
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
  • $0.6v$ to the left
  • B
    $1.4v$ to the left
  • C
    $v$ to the left
  • D
    $0.4v$ to the right

Answer

Correct option: A.
$0.6v$ to the left
a
since, $A$ and $B$ have same mass. So, after elastic collision, they interchange their velocity

$\therefore \mathrm{v}_{\mathrm{B}}=\mathrm{v}$

After collision between $\mathrm{B}$ and $\mathrm{C}$

$\mathrm{v}_{\mathrm{B}}=\left(\frac{\mathrm{m}-4 \mathrm{m}}{\mathrm{m}+4 \mathrm{m}}\right) \mathrm{v}+\left(\frac{2 \times 4 \mathrm{m} \times 0}{\mathrm{m}+4 \mathrm{m}}\right)=-\frac{3}{5} \mathrm{v}$

Again, collision takes place between $A$ and $B .$ So, velocity will be inter changed.

$\therefore v_{A}=\left(v_{B}\right)=\frac{3}{5} v$

in left ward direction.

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