MHT CET · Physics · Center of Mass Momentum and Collision
A body of mass 'm' moving with speed \(3 \mathrm{~m} / \mathrm{s}\) collides with a body of mass \({ }^{4} 2 \mathrm{~m}^{\prime}\) at rest. The coalesced mass will start to move with a speed of
- A \(3 \mathrm{~m} / \mathrm{s}\)
- B \(6 \mathrm{~m} / \mathrm{s}\)
- C \(9 \mathrm{~m} / \mathrm{s}\)
- D \(1 \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(D) \(1 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Initial velocity of moving mass \(\mathrm{u}_{1}=3 \mathrm{~km} / \mathrm{h}\)
Initial velocity of stationary mass \(\mathrm{u}_{2}=0\)
Thus momentum of the system initially \(P_{i}=m_{1} u_{1}+m_{2} u_{2}\)
\(\therefore P_{i}=3 m+0=3 m\)
Let the velocity of coalesced mass be V.
Thus momentum of the system finally \(\mathrm{P}_{\mathrm{f}}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{V}=3 \mathrm{mV}\)
Using conservation of linear momentum \(P_{i}=P_{f}\)
\(\therefore 3 \mathrm{~m}=3 \mathrm{~mV}\)
\(\Longrightarrow \mathrm{V}=1 \mathrm{~km} / \mathrm{h}\)
Initial velocity of stationary mass \(\mathrm{u}_{2}=0\)
Thus momentum of the system initially \(P_{i}=m_{1} u_{1}+m_{2} u_{2}\)
\(\therefore P_{i}=3 m+0=3 m\)
Let the velocity of coalesced mass be V.
Thus momentum of the system finally \(\mathrm{P}_{\mathrm{f}}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{V}=3 \mathrm{mV}\)
Using conservation of linear momentum \(P_{i}=P_{f}\)
\(\therefore 3 \mathrm{~m}=3 \mathrm{~mV}\)
\(\Longrightarrow \mathrm{V}=1 \mathrm{~km} / \mathrm{h}\)
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