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 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

  • 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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