Question 6.2: The Archer Goal Calculate recoil velocity using conservation...
The Archer Goal Calculate recoil velocity using conservation of momentum.
Problem An archer stands at rest on frictionless ice and fires a 0.500-kg arrow horizontally at 50.0 m/s. (See Fig. 6.7.) The combined mass of the archer and bow is 60.0 kg. With what velocity does the archer move across the ice after firing the arrow?
Strategy Set up the conservation of momentum equation in the horizontal direction and solve for the final velocity of the archer. The system of the archer (including the bow) and the arrow is not isolated, because the gravitational and normal forces act on it.
These forces, however, are perpendicular to the motion of the system and hence do no work on it.
Write the conservation of momentum equation. Let v_{1 f} \text { be the archer's velocity and } v_{2 f} the arrow’s velocity.
\begin{aligned} p_{i} &=p_{f} \\ 0 &=m_{1} v_{1 f}+m_{2} v_{2 f} \end{aligned}
Substitute m_{1}=60.0 kg , m_{2}=0.500 kg , and v_{2 f}=50.0 m / s , \text { and solve for } v_{1 f} :
\begin{aligned} v_{1 f} &=-\frac{m_{2}}{m_{1}} v_{2 f}=-\left(\frac{0.500 kg }{60.0 kg }\right)(50.0 m / s ) \\ &=-0.417 m / s \end{aligned}
Remarks The negative sign on v_{1 f} indicates that the archer is moving opposite the direction of motion of the arrow, in accordance with Newton’s third law. Because the archer is much more massive than the arrow, his acceleration and consequent velocity are much smaller than the acceleration and velocity of the arrow.
Newton’s second law, ∑F = ma, can’t be used in this problem because we have no information about the force on the arrow or its acceleration. An energy approach can’t be used either, because we don’t know how much work is done in pulling the string back or how much potential energy is stored in the bow. Conservation of momentum, however, readily solves the problem.