A small block of wood of inertia m_b is released from rest a distance h above th
ID: 1997016 • Letter: A
Question
A small block of wood of inertia m_b 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 m_p. 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 v_p. To what maximum height h_max 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, m_p, m_b, and v_p. How much energy is dissipated in the collection? Express your answer in terms of g and some of all of the variables h, d, m_p, m_b, and v_p.Explanation / Answer
A. In the collision only momentum is conserved
momentum before collision
pellet = mpvp
velocity of the block just before collision = (2gd)1/2
momentum of block before collision = -mb(2gd)1/2 , consider upward direction as +ve and downward as -ve
Total momentum before collision = mpvp - mb(2gd)1/2
let vf be te velocity of the combined block immediately after collision then conserving the momentum
(mp +mb)vf = mpvp - mb(2gd)1/2
maximum hieght reached by the block from the point of collision
y = vf2/2g = [(mpvp - mb(2gd)1/2 )/(mp +mb)]2/2g
height from ground = (h-d) +y = (h-d)+[(mpvp - mb(2gd)1/2 )/(mp +mb)]2/2g
KE before collison = mpvp2 /2 + mbgd
KE just after collision = (mp +mb)vf2 /2 = [mpvp - mb(2gd)1/2 ]2 /2(mp +mb)
energy dessipated in collission
= (mpvp2 /2 + mbgd) - [mpvp - mb(2gd)1/2 ]2 /2(mp +mb)