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Standard XII

Physics

The 'e' Equation

If two masses...

Question

If two masses $m_{1}$ and $m_{2}$ collide, the ratio of the changes in their respective velocities is proportional to:-

\frac{m_{1}}{m_{2}}

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\sqrt{\frac{m_{1}}{m_{2}}}

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\frac{m_{2}}{m_{1}}

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\sqrt{\frac{m_{2}}{m_{1}}}

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Solution

The correct option is B $\sqrt{\frac{m_{1}}{m_{2}}}$
$K E_{1} = K E_{2}$
$\frac{1}{2} m_{1} v_{1}^{2} = \frac{1}{2} m_{2} v_{2}^{2}$
$\frac{1}{2 m_{1}} {(m_{1} v_{1})}^{2} = \frac{1}{2 m_{2}} {(m_{2} v_{2})}^{2}$
$\Rightarrow \frac{1}{2 m_{1}} {(P A)}^{2} = \frac{1}{2 m_{2}} {(P B)}^{2}$
$\Rightarrow {(\frac{P A}{P B})}^{2} = \frac{m_{1}}{m_{2}}$
$\Rightarrow \frac{P A}{P B} = \sqrt{\frac{m_{1}}{m_{2}}} .$
Hence $,$ option $(B)$ is correct $.$

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If two masses m1 and m2 collide, the ratio of the changes in their respective velocities is proportional to:-

The correct option is B √m1m2KE1=KE212m1v21=12m2v2212m1(m1v1)2=12m2(m2v2)2⇒12m1(PA)2=12m2(PB)2⇒(PAPB)2=m1m2⇒PAPB=√m1m2.Hence, option (B) is correct.

If two masses $m_{1}$ and $m_{2}$ collide, the ratio of the changes in their respective velocities is proportional to:-