Two masses of 10 kg and 20 kg respectively are connected by a massless spring as shown in figure. A force of 200 N acts on the 20 kg mass at the instant when the 10 kg mass has an acceleration of $12$ towards right, the acceleration of the 20 kg mass is :

2 m s^{- 2}

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4 m s^{- 2}

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10 m s^{- 2}

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20 m s^{- 2}

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Solution

The correct option is B $4 m s^{- 2}$
$m_{1} = 10 k g; m_{2} = 20 k g; F_{2} = 200 N; a_{1} = 12 m s^{- 2}; a_{2} = ?$
For mass $m_{1}$ ; $F = m_{1} a_{1} = 10 \times 12 = 120 N$
For mass $m_{2}$ ;
$200 - F = 20 a_{2}$
$200 - 120 = 20 a_{2}$
$a_{2} = \frac{80}{20} = 4 m s^{- 2}$
hence,option B is correct.

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