Work Energy and Power - Result Question 54
57. A mass m moving horizontally (along the $x$-axis) with velocity $v$ collides and sticks to mass of $3 m$ moving vertically upward (along the $y$-axis) with velocity $2 v$. The final velocity of the combination is
[2011M]
(a) $\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j}$
(b) $\frac{1}{3} v \hat{i}+\frac{2}{3} v \hat{j}$
(c) $\frac{2}{3} v \hat{i}+\frac{1}{3} v \hat{j}$
(d) $\frac{3}{2} v \hat{i}+\frac{1}{4} v \hat{j}$
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Answer:
Correct Answer: 57. (a)
Solution:
- (a) As the two masses stick together after collision, hence it is inelastic collision. Therefore, only momentum is conserved.
$\therefore m v \hat{i}+3 m(2 v) \hat{j}=(4 m) \vec{v}$
$\vec{v}=\frac{v}{4} \hat{i}+\frac{6}{4} v \hat{j}$
$=\frac{v}{4} \hat{i}+\frac{3}{2} v \hat{j}$
