Example 9.1 The Archer
Let us consider the situation proposed at the beginning of
this section. 160kg archer stands at rest on frictionless ice
and fires a 0.50-kg arrow horizontally at 50 m s (Fig. 9.2).
With what velocity does the archer move across the ice after
firing the arrow​

v1f = -0.16 ms

Explanation:

Use the conservation law of linear momentum:

m1v1i + m2v2i = m1v1f + m2v2f

where

v1i = v2i = 0

m1 = 160 kg

m2 = 0.50 kg

v2f = 50m/s

v1f = ?

So we have

0 = (160 kg)v1f + (0.5 kg)(50 m/s)

v1f = -(25 kg-m/s)/(160 kg)

= -0.16 m/s

Note: the negative sign means that its direction is opposite that of the arrow.

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