A 57-kg archer on ice skates launches a 32.4-g arrow traveling 65 m/s.If the archer is initially at rest and can recoil without friction,the speed of the archer after the arrow launch is

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Quizplus · Accepted Answer

Momentum is conserved, so the total momentum before the launch is zero, and the total momentum after the launch is also zero. Therefore, the momentum of the archer after the launch is equal to the negative of the momentum of the arrow. The momentum of the arrow is:$$p_a = m_a v_a = (0.0324 	ext{ kg})(65 	ext{ m/s}) = 2.116 	ext{ kg m/s}$$Therefore, the momentum of the archer after the launch is:$$p_r = -p_a = -2.116 	ext{ kg m/s}$$The speed of the archer after the launch is:$$v_r = \frac{p_r}{m_r} = \frac{-2.116 	ext{ kg m/s}}{57 	ext{ kg}} = -0.0369 	ext{ m/s}$$The negative sign indicates that the archer is moving in the opposite direction of the arrow.

A 57-Kg Archer on Ice Skates Launches a 32