# 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?

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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.)

• 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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