Question

A 15g bullet is fired from a gun of mass 3 kg. If the speed of bullet just after firing is 300 ms^-1, what is the recoil speed of gun?

Answer 1

Step 1: Given data

Mass of bullet, $M_B=15g$

Mass of gun, $M_G=3kg$

$=3\times {10}^{3}g$ [Since, $1kg={10}^{3}g$]

Initial velocity of gun, $u_G=0m{s}^{-1}$

Initial velocity of bullet, $u_B=0m{s}^{-1}$

Final velocity of bullet, $v_B=300m{s}^{-1}$

Final velocity of gun (recoil speed), $v_G=?$

Step 2: Conserving the linear momentum

Law conservation of momentum states that in absence of external forces, the total momentum of a system remains constant.

So, the initial momentum of the gun-bullet system $P_i$ i.e. before firing must be equal to the final momentum of gun-bullet system $P_f$ i.e. after firing.

$P_i=P_f$

$M_G{u}_G+M_B{u}_B=M_G{v}_G+M_B{v}_B$

Step 3: Finding the recoil speed of gun

Substitute the values in the above equation.

$3\times {10}^3\times 0+15\times 0=3\times {10}^3\times v_G+15\times 300$

$0=3000\times v_G+4500$

$-4500=3000\times v_G$

$v_G=\frac{-4500}{3000}$

$v_G=\frac{-3}{2}$

$v_G=-1.5m{s}^{-1}$

The negative sign shows that the gun recoils in the opposite direction of the bullet velocity.

Hence, the recoil speed of the gun is $1.5m{s}^{-1}$.