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A loaded spring gun, initially at rest o...

A loaded spring gun, initially at rest on a horizontal frictionless surface fires a marble of mass `m` at an angle of elecation `theta`, then velocity of the gun just after the firing is `:`

A

`(mv_(0))/(M)`

B

`(mv_(0)cos theta)/(M)`

C

`(mv_(0)cos theta)/(M+m)`

D

`(mv_(0)cos 2 theta)/(M+m)`

Text Solution

Verified by Experts

The correct Answer is:
C
Muzzle velocity `=v_(m//g)=v_(0)`
Along `x-` direction `,`
`v_(m(x))=v_(g(x))=v_(0)cos theta`
By momentum conservation`:(M+m)(0)=m (v_(0)cos theta-v)-Mv`
`rArr v=(mv_(0)cos theta)/(M+m)`
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Knowledge Check

  • A loaded spring gun of mass M fires a bullet of mass m with a velocity v at an angle of elevation theta . The gun is initially at rest on a horizontal smooth surface. After firing, the centre of mass of the gun and bullet system

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