Two balls A and B of mass 0.10 kg and 0.25 kg respectively are connected by a stretched spring of negligible mass and placed on a smooth table when the balls are released simultaneously, the initial acceleration of ball B is 10 cm/s² westward. The magnitude and direction of acceleration of the ball A are

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$25 cm / \sec^{2}, westward$

$2 \cdot 5 cm / \sec^{2}, westward$

$25 cm / s^{2}, eastward$

$2 \cdot 5 cm / \sec^{2}, eastward$

answer is D.

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Detailed Solution

Here, the force provided by the spring is internal Hence p _s = constant

$or m_{A} v_{A} + m_{B} v_{B} = constant or \frac{d}{dt} [m_{A} v_{A} + m_{B} v_{B}] = 0 m_{A} a_{A} + m_{B} a_{B} = 0 (∵ \frac{d}{dt} V = a) ∴ a_{A} = - \frac{m_{B}}{m_{A}} \times a_{B} = - \frac{0 \cdot 25}{0 \cdot 10} \times 10 cm / \sec^{2} = - 25 cm / \sec^{2} (Westward) = 25 cm / \sec^{2} (Eastward)$

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