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Standard XII

Physics

Centripetal Acceleration

Mass m 1 hi...

Question

Mass $m_{1}$ hits and sticks with $m_{2}$ while sliding horizontally with velocity $v$ along the common line of centres of the three equal masses ( $m_{1} = m_{2} = m_{3} = m$ ). Initially masses $m_{2}$ and $m_{3}$ are stationary and the spring is unstretched. Minimum kinetic energy of $m_{2}$ is $y m v^{2} / 36$ . Find $y$ .

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Solution

Just after impact $v$ , combnation of $m_{1}, m_{2}$ has velocity $\frac{v}{2}$ .
At an instant the spring is at maximum compression, both blocks would be at rest.
From energy equation:
$\frac{1}{2} (2 m) ({\frac{v}{2}}^{2}) = \frac{1}{2} K x^{2}$
$\Rightarrow$ $x = \sqrt{\frac{m v^{2}}{2 k}}$
Since, the compresion of spring is $\frac{2}{3}$ of max of $m_{3}$ from centre of mass.
Max kinetic energy of $m_{3}$ $= \frac{1}{2} k (\frac{2}{3} x)^{2} = \frac{m v^{2}}{36}$
Therefore, $y = 1$

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Mass m1 hits and sticks with m2 while sliding horizontally with velocity v along the common line of centres of the three equal masses (m1=m2=m3=m). Initially masses m2 and m3 are stationary and the spring is unstretched. Minimum kinetic energy of m2 is ymv2/36. Find y.