Question: A bullet of mass 0.010 kg, moving horizontally, strikes a stationary b

Question

Question: A bullet of mass 0.010 kg, moving horizontally, strikes a stationary block of wood of mass 1.5 kg that is suspended as a pendulum. The bullet lodges in the wood, and together they swing upward a vertical distance of 0.40 m. a) For the block/bullet combination, calculate: i. the potential energy change at maximum height; ii. the velocity and the momentum just after the collision. b) Determine the velocity and the kinetic energy of the bullet before the collision c) How much energy is dissipated as heat during the collision? 0.010 kg 0.40 m d) If it takes a time of 5.0 ms for the bullet to imbed in the wood, what is the average force that the bullet exerts on the block? 1.5 kg

Explanation / Answer

a) i ) change in potential energy is dU = m*g*H = (0.01+1.5)*9.8*0.4 = 5.92 J

ii)

using law of conservation of energy

energy at lower point = energy at the height point

0.5*(m+M)*v^2 =(m+M)*g*H

0.5*v^2 = 9.8*0.4

v = 2.8 m/sec

momentum is p = (m+M)*v = (0.01+1.5)*2.8 = 4.228 Kg m/s

b)
using law of conservation of momentum

momentum after collision = momemtum before collision

4.228 = m*u

4.228 = 0.01*u

u = 4.228/0.01 = 422.8 m/sec

kinetic energy is KEi = 0.5*m*u^2 = 0.5*0.01*422.8^2 = 893.8 J

C) energy dissipated is (m+M)*g*H - (0.5*(m+M)*v^2

((0.01+1.5)*9.8*0.4) - (0.5*(0.01+1.5)*2.8^2) = 0 J

D) impulse = change in momentum

Favg*dt = m*(v-u)

Favg*(5*10^-3) = 0.01*(2.8-422.8) = -4.2

Favg = 4.2/(5*10^-3) = 840 N

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