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Question

# A bullet of mass m2 is fired from a gun of mass m1 with horizontal velocity v. A plane mirror is fixed to the gun facing towards the bullet. The velocity of the image of the bullet formed by the plane mirror with respect to the bullet is (Neglect the mass of mirror)

A
(1+m2m1)v
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B
(m1+m2m1)v
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C
2(m1+m2)vm1
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D
2(2m1+m2)vm1
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Solution

## The correct option is C 2(m1+m2)vm1Since mirror is mounted at the gun, the velocity of mirror (vm) and gun will be same Applying linear momentum conservation for bullet and gun system, m1 is mass of gun and mirror. Initial momentum of the system = 0. ⇒m1vm+m2(−v)=(m1+m2)×0 ⇒m1vm=m2v ∴vm=(m2m1)v Applying, (vo,m)⊥=−(vi,m)⊥ Taking rightwards +ve direction. ⇒(−v−vm)=−(vi−vm) or, −v−vm=−vi+vm or, vi=v+2vm ∴vi=v+2(m2m1)v=(m1+2m2m1)v Thus, the velocity of image w.r.t bullet (object) is, vi,o=vi−vo ⇒vi,o=vi−(−vo)=vi+vo or, vi,o=(m1+2m2m1)+v ∴vi,o=2(m1+m2)vm1

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