Question

A gun with inertia 4.5 kg fires a 12-g bullet at a stationary target located 1.0 km away. After the bullet leaves the gun, its speed decreases (constant x component of acceleration a = -1.0 m/s2 ) so that the bullet hits the target at 299 m/s.
If the direction in which the bullet moves is along a positive x coordinate axis, what is the recoil velocity of the gun?

Answer 1

As per the given question,

Mass of the gun$(M)=4.5 kg$

Mass of the bullet$(m)=12gm=12\times 10^{-3}kg$

Distance of the target $(d)=1k =1000 m$

deceleration of the bullets $a=-1 m/sec^2$

Hitting speed of the bullet to the target $(v)=299 m/sec$

Now, Let the initial speed of the bullet =u

So,

$v^2=u^2-2ad$

$\Rightarrow u^2=v^2+2ad$

$\Rightarrow v=\sqrt{299^2+2\times 1\times 1000}=302.32 m/sec$

Now applying the conservation of momentum,

Let the recoil speed of the gun $v_1$

$Mv_1+mu=0$

$\Rightarrow v_2=\dfrac{-mu}{ M}=\dfrac{-12\times 10^{-3}\times 302.32}{4.5}=-0.806 m/sec$

Here negative sign is representing that the bullet will move to negative x axis with the mentioned speed.