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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

a
2⋅5cm/sec2, westward
b
2⋅5cm/sec2, eastward
c
25cm/sec2, westward
d
25cm/s2, eastward

detailed solution

Correct option is D

Here, the force provided by the spring is internal Hence p s = constant  or mAvA+mBvB= constant   or ddtmAvA+mBvB=0 mAaA+mBaB=0 ∵ddtV=a ∴ aA=−mBmA×aB=−0⋅250⋅10×10cm/sec2 =−25cm/sec2 (Westward)  =25cm/sec2( Eastward )

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