Posted by **Annie** on Thursday, February 14, 2013 at 1:48am.

A 0.03-kg bullet is fired vertically at 200m/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 37m. (a). What was the speed of the baseball/bullet right after the collision? (b). What was the average force of air resistance while the baseball/bullet was rising?

- Physics -
**drwls**, Thursday, February 14, 2013 at 5:28am
(a) The actual speed V' of the bullet/ball combination after impact can be obtained by applying conservation of momentum.

m*200 = (m+M) V'

V' = (0.03/0.018)*200 = 33.3 m/s

If there were no air resistance, the ball and bullet would rise to a height H given by

(m+M)gH = (1/2)(m+M)V'^2

H = V'^2/(2g) = 56.7 m

Since it only rose 37 m, 19.7/56.7 = 34.7% of the ball/bullet's initial kinetic energy is lost due to air friction. That would be

(1/2)(m+M)*V'^2*(0.347)

= 34.6 Joules

Set that energy loss equal to Fav*37 m to get the average air friction force, Fav = 0.936 Newtons

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