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A single horizontal force F is applied to a block of mass $M_{1}$ , which is in contact with another block of mass $M_{2}$ , as in the below figure. If the surface is frictionless, then the force between the two blocks is__

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\frac{M_{1} F}{M_{2}}

\frac{M_{1} M_{2} F}{M_{1} + M_{2}}

\frac{M_{2} F}{M_{1} + M_{2}}

\frac{M_{2} F}{M_{1}}

answer is C.

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Detailed Solution

A single horizontal force F is applied to a block of mass

M_{1}

, which is in contact with another block of mass

M_{2}

as in the below figure. Given the surface is frictionless, the force between the block is

\frac{M_{2} F}{M_{1} + M_{2}} .

Given:
Two bodies of mass

M_{1}

and

M_{2}

Acceleration is same for both the bodies

= a

Let the force be

F

To find:
Force between the block
Since the acceleration is same and the resultant mass

m = M_{1} + M_{2} .

Now using,

F = ma

F = (M_{1} + M_{2}) a

Therefore,

a = \frac{F}{M_{1} + M_{2}}

----- (1)
Let

F_{1}

be the force between the blocks.

F_{1} = M_{2} \times a

[Mass

{= M}_{2}

because the force is applied on a block of mass

M_{1},

which is in contact with another block of mass

M_{2}]

{\Rightarrow F}_{1} = \frac{M_{2} F}{M_{1} + M_{2}}

[using equation (1)]

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