Need help with Physics!!!!! Any help is greatly appreciated.?

Question

A 0.04 kg bullet is fired vertically at 190 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. (a) (b) What was the average force of air resistance while the baseball/bullet was rising? (Consider the positive direction to be upwards.)
What was the speed of the baseball/bullet right after the collision?

Answer 1

This problem involves two principles: conservation of momentum and conservation of total energy. The collision of bullet and baseball is inelastic, conservation of momentum applies. Just before the collision,the momentum of the bullet is m(bu)*v(bu), and the baseball's momentum is zero. After the collision, the combined momentum is [m(bb)+m(bu)]*v(bb). So you can find the velocty of the baseball/bullet right after collision from:

m(bu)*v(bu) = [m(bb)+m(bu)]*v(bb)

v(bb) = m(bu)*v(bu)/ [m(bb)+m(bu)].

This is the answer to the second part of your question.

The initial kinetic energy (KE) of the combined masses is .5*[m(bb)+m(bu)]*v(bb)^2; this energy is propels the masses to a height h above the imapact point. The combined masses have gained a potential energy (PE) of [m(bb)+m(bu)]*g*h. This potential energy must equal the initial kinetic energy less the energy lost to air drag. The latter is given by ∫F*dy, where F is the air friction force. This integral is equal to F(avg)*h. So summing the total energy balance

PE(final) = KE(initial) - air friction energy loss

or

air friction loss = KE(initial) - PE(final)

Substituting these quantities:

F(avg)*h = .5*[m(bb)+m(bu)]*v(bb)^2 - [m(bb)+m(bu)]*g*h

F(avg) = {.5*[m(bb)+m(bu)]*v(bb)^2}/h - [m(bb)+m(bu)]*g

Use v(bb) calculated above.