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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a
2⋅5cm/sec2, westward
b
2⋅5cm/sec2, eastward
c
25cm/sec2, westward
d
25cm/s2, eastward
answer is D.
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Detailed Solution
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 )