question_answer

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 2v. The final velocity of the combination is     [AIPMT (M) 2011]

A) \[\frac{1}{4}v\mathbf{\hat{i}}+\frac{3}{2}v\mathbf{\hat{j}}\]

B) \[\frac{1}{3}v\mathbf{\hat{i}}+\frac{2}{3}v\mathbf{\hat{j}}\]

C) \[\frac{2}{3}v\mathbf{\hat{i}}+\frac{1}{3}v\mathbf{\hat{j}}\]

D) \[\frac{3}{2}v\mathbf{\hat{i}}+\frac{1}{4}v\mathbf{\hat{j}}\]

Correct Answer: A

Solution :

From the law of conservation of linear momentum

\[mv+(3m)(2v)=(4m)v'\]

\[mv\ \mathbf{i}+6\,mv\,\mathbf{j}=4\,mv'\]

\[v'=\frac{1}{4}v\,\mathbf{i}+\frac{6}{4}v\,\mathbf{j}\]

\[v'=\frac{1}{4}v\,\mathbf{i}+\frac{3}{2}v\,\mathbf{j}\]