Answer
(a) v = 4.32 m/s
(b) When the skate boarder and the sack meet, there is a perfectly inelastic collision. The sack exerts an impulse on the skateboarder and the skateboarder exerts an impulse on the sack. After the collision, they move together at the same speed and the horizontal momentum in the system is conserved.
(c) The skateboarder's speed will still be 4.32 m/s.
Work Step by Step
(a) $p = (60~kg)(4.5~m/s) = 270~kg~m/s$
$mv = 270~kg~m/s$
$v = \frac{270~kg~m/s}{m} = \frac{270~kg~m/s}{62.5~kg}$
$v = 4.32~m/s$
(b) When the skate boarder and the sack meet, there is a perfectly inelastic collision. The sack exerts an impulse on the skateboarder and the skateboarder exerts an impulse on the sack. After the collision, they move together at the same speed and the horizontal momentum in the system is conserved.
(c) If the sack is thrown straight up from the skateboarder's point of view, the sack's horizontal motion will not change. Therefore, because of conservation of the horizontal component of momentum, the skateboarder's speed will still be 4.32 m/s.
