A 0.025-kg bullet is fired vertically at 208 m/s into a 0.15-kg baseball that is initially at rest. The bullet lodges in the baseball and, after the collision, the baseball/bullet rise to a height of 37 m.
What was the average force of air resistance while the baseball/bullet was rising? (Consider the positive direction to be upwards.)
1 Answer
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Assume momentum is conserved in the collision and solve for the initial velocity of the bullet-ball…
MV (bullet) = MV (total)
(0.025)(208) = (0.025 + 0.15)V
V = 29.71 m/s
Use conservation of energy to find the height the bullet-ball would have risen without air resistance…
½mv² = mgh
h = v² / 2g
h = (29.71)² / (2)(9.81)
h = 45.0 m
The difference in potential energy between the ideal height and the actual height is the amount of energy lost to work against air resistance….
mgh2 – mgh1 = FΔh
(0.175)(9.81)(45) – (0.175)(9.81)(37) = F(37)
F = 0.371 N <=========================
