A small block of wood of inertia mb is released from rest a distance h above the

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)

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