# A rifle with a weight of 30 N fires a 5.0 g bullet with a speed of 300m/s. (a) Find the recoil speed of the rifle. (b) If a700 N man holds the rilfe firmly against his shoulder, find there coil speed of the man and his rifle.

A rifle with a weight of 30 N fires a 5.0 g bullet with a speed of 300m/s. (a) Find the recoil speed of the rifle. (b) If a700 N man holds the rilfe firmly against his shoulder, find there coil speed of the man and his rifle.

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Nathanael Webber
${m}_{1}{u}_{1}+{m}_{2}{u}_{2}={m}_{1}{v}_{1}+{m}_{2}{v}_{2}$
initially both gun and bullet are at rest
${m}_{1}{v}_{1}+{m}_{2}{v}_{2}=0$
${v}_{1}=-\frac{{m}_{2}{v}_{2}}{{m}_{1}}$

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Jeffrey Jordon

The change in the momentum of an isolated system is zero where the initial momentum equals the final momentum. The particles interact in the system without interact with the environment. So, the momentum is conserved and we use the law of conservation of momentum as given by equation (6.7) in the form

${p}_{f}={p}_{i}$

The system includes the riﬂe and the bullet. We are given the weight of the rifle by w= 30 N. So, we can calculate its mass as next

Initially, the bullet and the rifle are at rest, so their initial velocities are zero. Hence, the initial momentum of the system is zero

${p}_{f}=0$

${m}_{r}{v}_{r}+{m}_{b}{v}_{b}=0$

${v}_{r}=\frac{-{m}_{b}{v}_{b}}{{m}_{r}}$

Now, we plug the values for into equation (2) to get ${v}_{r}$

${v}_{r}=\frac{-{m}_{b}{v}_{b}}{{m}_{r}}$

=- 0.49 m/s

The velocity of the rifle is  in the opposite direction of the bullet.