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Energy in Collision

Two blocks of...

Question

Two blocks of mass m₁ and m₂ interconnected with a spring of stiffness 'k' are kept on a smooth horizontal surface. Which of the following ratios are correct, if the spring was extended and released?

(Where X = displacement, v = speed, P = momentum, a = acceleration, all relative to the surface, KE = kinetic energy, F = force.)

A

$\frac{F_{1}}{F_{2}} = 1 & \frac{X_{1}}{X_{2}} = 1$

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B

$\frac{v_{1}}{v_{2}} = \frac{m_{2}}{m_{1}} & \frac{K E_{1}}{K E_{2}} = \frac{m_{2}}{m_{1}}$

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C

$\frac{P_{1}}{P_{2}} = 1 & \frac{K E_{1}}{K E_{2}} = \frac{m_{1}}{m_{2}}$

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D

$\frac{X_{1}}{X_{2}} = \frac{m_{2}}{m_{1}} & \frac{a_{1}}{a_{2}} = \frac{m_{1}}{m_{2}}$

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Solution

The correct option is B
$\frac{v_{1}}{v_{2}} = \frac{m_{2}}{m_{1}} & \frac{K E_{1}}{K E_{2}} = \frac{m_{2}}{m_{1}}$

$(F_{e x t})_{s y s t e m} = 0 \Rightarrow {\to P}_{1} + {\to P}_{2} = 0 \Rightarrow p_{1} = p_{2} . \Rightarrow m_{1} {\to v}_{1} + m_{2} {\to v}_{2} = 0 \Rightarrow \frac{v_{1}}{v_{2}} = \frac{m_{2}}{m_{1}} \frac{K E_{1}}{K E_{2}} = \frac{\frac{p_{1}^{2}}{2 m_{1}}}{\frac{p_{2}^{2}}{2 m_{2}}} = \frac{m_{2}}{m_{1}} (s i n c e p_{1} = p_{2}) \frac{F_{1}}{F_{2}} = \frac{k x}{k x} W h e r e, x = d e f o r m a t i o n o f t h e s p r i n g \Rightarrow \frac{F_{1}}{F_{2}} = 1 \Rightarrow \frac{m_{1} a_{1}}{m_{2} a_{2}} = 1$

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{m_{2}}{m_{1}}$

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