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Q.

A spring connects two particles of masses $m_{1} and m_{2}$ . A horizontal force F acts on $m_{1}$ . Ignoring friction, when the elongation of the spring is x then :

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a

a_{1} = \frac{F - k x}{m_{1}}

b

F = m_{1} a_{1} + m_{2} a_{2}

c

a_{2} = \frac{k x}{m_{2}}

d

a_{1} = a_{2} = \frac{F}{m_{1} + m_{2}}

at the time of maximum elongation of the spring

answer is A, B, C, D.

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Detailed Solution

F - K x = m_{1} a_{1}

$a_{1} = \frac{F - K x}{m_{1}}$

$K x = m_{2} a_{2}$

$a_{2} = \frac{K x}{m_{2}}$

$F = m_{1} a_{1} + m_{2} a_{2}$

$a_{1} = a_{2} = \frac{F}{m_{1} + m_{2}}$

At the time of maximum elongation both blocks attain same acceleration

So

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