Posted by Annie on .
A 0.03kg bullet is fired vertically at 200m/s into a 0.15kg 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,
(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