Question

A $50\mathrm{kg}$archer, standing on frictionless ice, shoots a $100\mathrm{g}$ arrow at a speed of $100\mathrm{m}/\mathrm{s}$. What is the recoil speed of the archer?

Short answer

The recoil of the archer is$0.2\mathrm{m}/\mathrm{s}.$

Step-by-step solution

Given information

we have given,

$$\begin{gathered} \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{archer}=50\mathrm{kg} \\ \mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{arrow}=100\mathrm{g} \\ \mathrm{Speed}\mathrm{of}\mathrm{the}\mathrm{arrow}=100\mathrm{m}/\mathrm{s} \end{gathered}$$

we have to find the recoil speed of the archer.

Simplify

When two particles collide and stick together and move as one body then it will be inelastic collision where only momentum will be conserved.

Since initially the archer and the arrow is at rest then the initial momentum and final momentum will be zero.

It means that the magnitude of momentum of archer will be equal to the magnitude of the momentum of arrow and opposite in direction.

Then,

$$\begin{gathered} {\mathrm{m}}_{\mathrm{arrow}}{\mathrm{v}}_{\mathrm{arrow}}=-{\mathrm{m}}_{\mathrm{archer}}{\mathrm{v}}_{\mathrm{archer}} \\ {\mathrm{v}}_{\mathrm{archer}}=\frac{-{\mathrm{m}}_{\mathrm{arrow}}{\mathrm{v}}_{\mathrm{arrow}}}{{\mathrm{m}}_{\mathrm{archer}}} \\ {\mathrm{v}}_{\mathrm{archer}}=\frac{-0.1\mathrm{kg}\times 100\mathrm{m}/\mathrm{s}}{50\mathrm{kg}} \\ {\mathrm{v}}_{\mathrm{archer}}=-0.2\mathrm{m}/\mathrm{s} \end{gathered}$$