Question

# Two blocks with masses m1=3 kg and m2=4 kg are touching each other on a fricionless surface. If a force of 5 N is applied on m1 as shown in figure. The acceleration of each block is

A
a1=57 m/s2,a2=54 m/s2
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B
a1=53 m/s2,a2=54 m/s2
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C
a1=54 m/s2,a2=53 m/s2
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D
a1=57 m/s2,a2=57 m/s2
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Solution

## The correct option is D a1=57 m/s2,a2=57 m/s2By the FBD of m1: 5−R=m1a ...(i) where R is the reaction on the block m1 by the block m2 By the FBD of m2: R=m2a ....(ii) From (i) and (ii) we get and R1 and R2 does not contribute to horizontal accelerations. 5=(m1+m2)a ⇒a=53+4=57 m/s2 both the blocks move together with the same acceleration Hence option D is the correct answer

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