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A 30 g bullet is fired from a 1.6 kg rifle at a target. If the muzzle velocity of the bullet is 360 m/s, what is the recoil velocity of the rifle?

Answer :

Answer:

Approximately [tex]6.75\; {\rm m\cdot s^{-1}}[/tex].

Explanation:

Let [tex]u[/tex] denote the initial velocity and let [tex]v[/tex] denote the velocity after launching.

By the conservation of momentum, the sum of momentum would the same before and after launching:

[tex]m_{b}\, u_{b} + m_{r} \, u_{r} = m_{b}\, v_{b} + m_{r}\, v_{r}[/tex].

Assuming that [tex]u_{b} = u_{r} = 0\; {\rm m\cdot s^{-1}}[/tex]:

[tex]m_{b}\, v_{b} + m_{r}\, v_{r} = 0[/tex].

It is given that [tex]v_{b} = 360\; {\rm m\cdot s^{-1}}[/tex] and [tex]m_{r} = 1.6\; {\rm kg}[/tex]. Apply unit conversion and ensure that mass values are measured in the same unit (kilograms):

[tex]m_{b} = 30\; {\rm g} = 30 \times 10^{-3}\; {\rm kg} = 0.030\; {\rm kg}[/tex].

Substitute these values into the equation and solve for [tex]v_{r}[/tex]:

[tex]\begin{aligned}v_{r} &= \frac{-m_{b}\, v_{b}}{m_{r}}\\ &= \frac{-(0.030\; {\rm kg})\, (360\; {\rm m\cdot s^{-1}})}{1.6\; {\rm kg}} \\ &= 6.75\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

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Rewritten by : Jeany

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