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Question: Two balls of masses \(m _ { 1 }\) and \(m _ { 2 }\) are separated from each other by a powder ch...

Two balls of masses m1m _ { 1 } and m2m _ { 2 } are separated from each other by a powder charge placed between them. The whole system is at rest on the ground. Suddenly the powder charge explodes and masses are pushed apart. The mass m1m _ { 1 } travels a distance S1S _ { 1 } and stops. If the coefficients of friction between the balls and ground are same, the mass m2m _ { 2 } stops after travelling the distance

A

s2=m1m2s1s _ { 2 } = \frac { m _ { 1 } } { m _ { 2 } } s _ { 1 }

B

s2=m2m1s1s _ { 2 } = \frac { m _ { 2 } } { m _ { 1 } } s _ { 1 }

C

s2=m12m22s1s _ { 2 } = \frac { m _ { 1 } ^ { 2 } } { m _ { 2 } ^ { 2 } } s _ { 1 }

D

s2=m22m12s1s _ { 2 } = \frac { m _ { 2 } ^ { 2 } } { m _ { 1 } ^ { 2 } } s _ { 1 }

Answer

s2=m12m22s1s _ { 2 } = \frac { m _ { 1 } ^ { 2 } } { m _ { 2 } ^ { 2 } } s _ { 1 }

Explanation

Solution

We know that in the given condition s1m2s \propto \frac { 1 } { m ^ { 2 } }

\therefore s2s1=(m1m2)2\frac { s _ { 2 } } { s _ { 1 } } = \left( \frac { m _ { 1 } } { m _ { 2 } } \right) ^ { 2 }s2=(m1m2)2×s1s _ { 2 } = \left( \frac { m _ { 1 } } { m _ { 2 } } \right) ^ { 2 } \times s _ { 1 }