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Question

# Two particles of mass 1 kg and 2 kg are placed at a separation of 50 cm. Assuming that the only forces acting on the particles is due to their mutual attraction, find the initial acceleration of the two particles.

A
5.3×1010 m/s2, 2.65×1010 m/s2
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B
4.3×1010 m/s2, 2.65×1010 m/s2
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C
6.3×1010 m/s2, 2.65×1010 m/s2
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D
2.3×1010 m/s2, 2.25×1010 m/s2
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Solution

## The correct option is A 5.3×10−10 m/s2, 2.65×10−10 m/s2 The force due to mutual attraction exerted on one particle by the another one is F=Gm1m2R2 F=5.3×10−10 N Now we have to calculate acceleration of 1 kg a1=Fm1=5.3×10−10 m/s2 The acceleration is directed towards the 2 kg particle. The acceleration of 2 kg particle is a2=Fm2=2.65×10−10 m/s2 The acceleration of 2 kg particle will be directed towards the 1 kg particle.

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