A moving body with a mass m 1 strikes a stationary body of mass m 2- The masses of m 1 and m 2 should be in the ration m 1 m 2 so as to decrease the velocity of the first body 1-5 times assuming a perfectly elastic impact- Then the ratio m 1 m 2 is

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Standard VII

Physics

Law of Conservation of Energy

A moving body...

Question

A moving body with a mass $m_{1}$ strikes a stationary body of mass $m_{2}$ . The masses of $m_{1}$ and $m_{2}$ should be in the ration $\frac{m_{1}}{m_{2}}$ so as to decrease the velocity of the first body $1.5$ times assuming a perfectly elastic impact. Then the ratio $\frac{m_{1}}{m_{2}}$ is

\frac{1}{25}

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\frac{1}{5}

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5

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25

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Solution

The correct option is C $5$
Here, coefficient of restitution, $e$ is $1$ .
Let initial velocity of mass $m_{1}$ be $v_{1}$ and final velocity of mass $m_{2}$ be $v_{2}$ . According to the question, the final velocity of mass $m_{1}$ is $v_{1} / 1.5 = 2 v_{1} / 3$ .
So, $e = 1 = (v_{2} - 2 v_{1} / 3) / v_{1}$
$\Rightarrow v_{2} = 5 v_{1} / 3 - (1)$
Now by conservation of momentum, $m_{1} v_{1} = m_{1} v_{1} / 1.5 + m_{2} v_{2}$
using $(1)$ , $5 m_{2} v_{1} / 3 = m_{1} v_{1} / 3$
$\Rightarrow m_{1} / m_{2} = 5$

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A moving body with a mass m1 strikes a stationary body of mass m2. The masses of m1 and m2 should be in the ration m1m2 so as to decrease the velocity of the first body 1.5 times assuming a perfectly elastic impact. Then the ratio m1m2 is

A moving body with a mass $m_{1}$ strikes a stationary body of mass $m_{2}$ . The masses of $m_{1}$ and $m_{2}$ should be in the ration $\frac{m_{1}}{m_{2}}$ so as to decrease the velocity of the first body $1.5$ times assuming a perfectly elastic impact. Then the ratio $\frac{m_{1}}{m_{2}}$ is