question_answer

A mass m moving horizontally (along the x-axis) with velocity v collides and sticks to mass of 3m moving vertically upward (along the y-axis) with velocity 2v. The final velocity of the combination is

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 \[mv+(3m)(2v)=(4m)v'\] \[mv\ \mathbf{i}+6\,mv\,\mathbf{j}=4\,mv'\]                 \[v'=\frac{1}{4}v\,\mathbf{i}+\frac{6}{4}v\,\mathbf{j}\]                 \[v'=\frac{1}{4}v\,\mathbf{i}+\frac{3}{2}v\,\mathbf{j}\]