MHT CET · Physics · Center of Mass Momentum and Collision
A mass ' \(M\) ' moving with velocity ' \(\mathrm{V}\) ' along \(\mathrm{x}\)-axis collides and sticks to another mass \(2 \mathrm{M}\) which is moving along \(\mathrm{Y}\)-axis with velocity ' \(3 \mathrm{~V}\) '. The velocity of the combination after collision is
- A \(\frac{V}{3} \hat{i}+2 V \hat{j}\)
- B \(\frac{V}{2} \hat{\mathrm{i}}+\mathrm{V} \hat{\mathrm{j}}\)
- C \(\frac{V}{3} \hat{\mathrm{i}}-2 \mathrm{~V} \hat{\mathrm{j}}\)
- D \(\frac{V}{2} \hat{\mathrm{i}}-\mathrm{V} \hat{\mathrm{j}}\)
Answer & Solution
Correct Answer
(A) \(\frac{V}{3} \hat{i}+2 V \hat{j}\)
Step-by-step Solution
Detailed explanation
From law of conservation of linear momentum,
\(\begin{aligned}
& M v \hat{i}+2 M(3 v \hat{j})=3 M \vec{v} \\
& v \hat{i}+6 v \hat{j}=3 \vec{v} \\
& \vec{v}=\frac{v}{3} \hat{i}+2 v \hat{j}
\end{aligned}\)
\(\begin{aligned}
& M v \hat{i}+2 M(3 v \hat{j})=3 M \vec{v} \\
& v \hat{i}+6 v \hat{j}=3 \vec{v} \\
& \vec{v}=\frac{v}{3} \hat{i}+2 v \hat{j}
\end{aligned}\)
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