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?
A 0.025-kg bullet is fired vertically at a speed of 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 and bullet rise to a height of 37 meters. What was the average force of air resistance acting on the baseball and bullet during their ascent? (For this analysis, consider the positive direction to be upwards.)
1 Answers
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 <=========================
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