A small block of wood of inertia mb is released from rest a distance h above the
ID: 1504349 • Letter: A
Question
A small block of wood of inertia mb is released from rest a distance h above the ground, directly above your head. You decide to shoot it with your pellet gun, which fires a pellet of inertia mp. After the block has fallen a distance d, the pellet hits it and becomes embedded in it, kicking it upward. At the instant of impact, the pellet is moving at speed vp.
Part A
To what maximum height hmax does the pellet-block system rise above the ground?
Express your answer in terms of g and some or all of the variables h, d, mp, mb, and vp.
Part B
How much energy is dissipated in the collision?
Express your answer in terms of g and some or all of the variables h, d, mp, mb, and vp.
Explanation / Answer
A. speed of block after falling down distance d.
Applying vf^2 - vi^2 = 2ad
u^2 - 0 =2 g d
u = sqrt(2gd) ........downward
Applying momentum conservation for collision,
mb(-u) + mp vp = (mp + mb)v
v = (mb sqrt(2gd) + mp vp ) / (mp + mb)
height from collision point,
0^2 - v^2 = 2 g H
H = ((mb sqrt(2gd) + mp vp ) / (mp + mb) )^2 / 2g
hmax = H + (h - d)
= (2 g d mb^2 + mp^2 vp^2 + 2 mp mb vp sqrt(2gd) + 2gh mp^2 + 2gh mb^2 - 2gd mp^2 - 2gd mb^2 - 2mp mb d + 2mp mb ) / (2g (mp + mb)^2 )
= mp^2 vp^2 + 2 mp mb vp sqrt(2gd) + 2gh mp^2 + 2gh mb^2 - 2gd mp^2 - 2mp mb d + 2mp mb ) / (2g (mp + mb)^2 )
B) energy lost = mp vp^2 /2 + mb u^2 /2 - (mp + mb) v^2 / 2
= mp vp^2 /2 + g d mb - [ (mb + mp) ( (mb sqrt(2gd) + mp vp ) / (mp + mb) )^2 /2 ]
= mp vp^2 /2 + g d mb - [ (mp^2 vp^2 + 2 g d mb + 2 mb mp vp sqrt(2gd)) / 2(mp + mb) ]
= ( mp^2 vp^2 + mp mb vp^2 + 2g d mb mp + 2 g d mb^2 - mp^2 vp^2 - 2 g d mb - 2 mb mp vp sqrt(2gd) ) / 2 (mp + mb)
= ( mp mb vp^2 + 2g d mb mp + 2 g d mb^2 - 2 g d mb - 2 mb mp vp sqrt(2gd) ) / 2 (mp + mb)