MHT CET · Physics · Center of Mass Momentum and Collision
In a system of two particles of masses 'mi' and 'm2', the first particle is moved by a distance 'd' towards the centre of mass. To keep the centre of mass unchanged,
the second particle will have to be moved by a distance
- A \(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \mathrm{~d}\), towards the centre of mass.
- B \(\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}} \mathrm{~d}\), away from the centre of mass.
- C \(\frac{\mathrm{m}_{2}}{\mathrm{~m}_{1}} \mathrm{~d}\), towards the centre of mass.
- D \(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \mathrm{~d}\), away from the centre of mass.
Answer & Solution
Correct Answer
(A) \(\frac{\mathrm{m}_{1}}{\mathrm{~m}_{2}} \mathrm{~d}\), towards the centre of mass.
Step-by-step Solution
Detailed explanation
the 2 masses are miand m2
let \(x\) and \(y\) be the distance of miand \(m 2\) from the centre of mass respectively...
now, \(m_{1} x=m_{2} y\)
the mass \(m 1\) is moved by a distance \(d\),
let the mass \(\mathrm{m} 2\) be moved by a distance \(\mathrm{D}\)
therefore, \(m_{1}(x-d)=m_{2}(y-D)\)
\(\mathrm{m}_{1} \mathrm{x}-\mathrm{m}_{1} \mathrm{~d}=\mathrm{m}_{2} \mathrm{y}-\mathrm{m}_{2} \mathrm{D}\)
by eqn (i) \(m_{1} x=m_{2} y\)
\(=>-\mathrm{m}_{1} \mathrm{~d}=-\mathrm{m}_{2} \mathrm{D}\)
\(=\mathrm{D}=\frac{\mathrm{m}_{1} \mathrm{~d}}{\mathrm{~m} 2}\)
let \(x\) and \(y\) be the distance of miand \(m 2\) from the centre of mass respectively...
now, \(m_{1} x=m_{2} y\)
the mass \(m 1\) is moved by a distance \(d\),
let the mass \(\mathrm{m} 2\) be moved by a distance \(\mathrm{D}\)
therefore, \(m_{1}(x-d)=m_{2}(y-D)\)
\(\mathrm{m}_{1} \mathrm{x}-\mathrm{m}_{1} \mathrm{~d}=\mathrm{m}_{2} \mathrm{y}-\mathrm{m}_{2} \mathrm{D}\)
by eqn (i) \(m_{1} x=m_{2} y\)
\(=>-\mathrm{m}_{1} \mathrm{~d}=-\mathrm{m}_{2} \mathrm{D}\)
\(=\mathrm{D}=\frac{\mathrm{m}_{1} \mathrm{~d}}{\mathrm{~m} 2}\)
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