Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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

$\frac{3}{2}v\hat i+\frac{1}{4}v\hat j$

$\frac{1}{4}v\hat i+\frac{3}{2} v\hat j$

$\frac{1}{3}v\hat i+\frac{2}{3} v\hat j$

$\frac{2}{3} v\hat i+\frac{1}{3}v\hat j$

Step-by-Step Solution

Identify the System: This is a completely inelastic collision in two dimensions where the two masses stick together. Linear momentum is conserved.
Initial Momentum:

Mass 1 ( $m$ ): $\vec{p}_1 = m v \hat{i}$
Mass 2 ( $3m$ ): $\vec{p}_2 = 3m (2v) \hat{j} = 6mv \hat{j}$
Total Initial Momentum: $\vec{P}_{initial} = mv \hat{i} + 6mv \hat{j}$

Final Momentum:

Combined Mass: $M = m + 3m = 4m$
Final Velocity: $\vec{v}_f$
Total Final Momentum: $\vec{P}_{final} = (4m) \vec{v}_f$

Apply Conservation of Momentum: $\vec{P}_{initial} = \vec{P}_{final}$ $mv \hat{i} + 6mv \hat{j} = 4m \vec{v}_f$ $\vec{v}_f = \frac{mv \hat{i} + 6mv \hat{j}}{4m} = \frac{1}{4}v \hat{i} + \frac{6}{4}v \hat{j} = \frac{1}{4}v \hat{i} + \frac{3}{2}v \hat{j}$ [Class 11 Physics, Ch 6, Sec 5.11.3]

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