Rotational Motion Question 115
Three blocks are initially placed . 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 blocks $A$ is
Options:
A) 0.6$v$ to the left
B) 1.4$v$ to the left
C) $\vec{v}$ to the left
D) 0.4 $v$ to the right
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Answer:
Correct Answer: A
Solution:
Since, $A$ and $^{12}\text{C}$ have the same mass.
So, after elastic collision, they interchange their velocity.
$\therefore v_B=v$
After collision between $B$ and $C$, the system’s total momentum is conserved. $ v_B^{\prime}=(\frac{m-4 m}{m+4 m}) v+(\frac{2 \times 4 m \times v}{m+4 m})=-\frac{3}{5} v $
Again, collision takes place between $\mathrm{A}$ and $B$.
So, velocity will be exchanged.
$\therefore v_A^{\prime}=v_B^{\prime}=\frac{3}{5} v$ toward left
BETA
