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Question

Two blocks A and B of mass mA and mB respectively are kept in contact on a frictionless table. The experimenter pushes the block A from behind so that the block B, what is the force exerted by the experimenter on A ?

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Solution

Let , F = contact force between mA and mB

and f = force exerted by experimenter

F=mAaf=0

F=fmAa .....(i)

Now, mB.aF=0

F=mBa .....(ii)

From equation (i) and equation (ii), we have

fmAa=mBa

f=mBa+mAa

=a(mB+mA)

f=Fmb(mB+mA)

=F(1+mAmB)

[since a=FmB]

Force exerted by the experimenter

= =F(1+mAmB)


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