3. A rubber bullet of mass \"m\" and collides with a hanging pendulum bob of len

Question

3. A rubber bullet of mass "m" and collides with a hanging pendulum bob of length "L" and mass "6m". The head on collision occurs with perfect clasticity. After the collision the pendulum swings in a vertical circle and ust makes it over the top without falling inward. Given [L] Determine: a. The speed of the pendulum bob at the top of the loop. b. The speed of the pendulum bob just after the collision. c. The speed of the bullet just after the collision. d. The speed of the bullet before the collision. om

Explanation / Answer

a)

vtop = speed at the top = sqrt(gL)

b)

vbottom = speed at the bottom for looping the loop = sqrt(5gL)

vpf = speed of pendulum just after collision = vbottom = sqrt(5gL)

c)

vbf = velocity of bullet just after collision

vbi = velocity of bullet just before collision

using conservation of momentum

mb vbi + mp vpi = mb vbf + mp vpf

(m)vbi + (6m) (0) = (m) vbf + (6m) (sqrt(5gL))

vbi = vbf + 6sqrt(5gL) eq-1

using conservation of kinetic energy

mb v2bi + mp v2pi = mb v2bf + mp v2pf

(m)v2bi + (6m) (0)2 = (m) v2bf + (6m) (sqrt(5gL))2

(m) (vbf + 6sqrt(5gL))2 + (6m) (0)2 = (m) v2bf + (30 mgL)

(vbf + 6sqrt(5gL))2 = v2bf + 30gL

v2bf + 12 vbf sqrt(5gL) + 180gL = v2bf + 30gL

12 vbf sqrt(5gL) = - 150 gL

vbf = -5.6 sqrt(gL)

using eq-1

vbi = vbf + 6sqrt(5gL)

vbi = -5.6 sqrt(gL) + 6sqrt(5gL) = 7.8 sqrt(gL)

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