A gun with mass M fired a bullet having mass m with speed v relative to barrel of the gun which is inclined at an angle of with horizontal. The gun is placed over a smooth horizontal surface find the recoil speed of the gun.

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Solution

Let, M= mass of the gun
m= mass of the bullet
According to conservation of momentum,
$M v c o s θ + m v^{/} = 0$ [ $θ$ = angle with horizontal at which bullet is fired]

Therefore, recoil velocity, $v^{/} = \frac{M v c o s θ}{m}$

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A bullet of mass m leaves a gun of mass M kept on a smooth horizontal surface. If speed of the bullet relative to the gun is v, then recoil speed of the gun will be:

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A gun with mass M fired a bullet having mass m with speed v relative to barrel of the gun which is inclined at an angle of with horizontal. The gun is placed over a smooth horizontal surface find the recoil speed of the gun.

Let, M= mass of the gun m= mass of the bulletAccording to conservation of momentum,Mvcosθ+mv/=0 [θ= angle with horizontal at which bullet is fired]Therefore, recoil velocity, v/=Mvcosθm

Let, M= mass of the gun
m= mass of the bullet
According to conservation of momentum,
$M v c o s θ + m v^{/} = 0$ [ $θ$ = angle with horizontal at which bullet is fired]

Therefore, recoil velocity, $v^{/} = \frac{M v c o s θ}{m}$