A bullet of mass 0.010 kg, moving horizontally, strikes a block of wood of mass

Question

A bullet of mass 0.010 kg, moving horizontally, strikes a block of wood of mass 1.5 kg that is suspended as pendulum. The bullet lodges in the wood, and together they swing upward a vertical distance of 0.40 m. a) Calculate the kinetic energy and hence the velocity of the block/bullet combination just after the collision. b) Determine the velocity of the bullet before the collision. c) If it takes 5.0 milliseconds for the bullet to imbed in the wood, calculate the average force that the bullet exerts on the block. Bonus part: How much energy is dissipated as sound and heat during the collision?

Explanation / Answer

mass of bullet = m= 0.010 kg,

mass of block = M= 1.5 kg

gain in height = 0.40 m

a) conserving energy,

( m+M) gh = 1/2 ( m+M) v^2

PE = KE

( 1.51) (9.8 ) (0.40 )= 0.5( 1.51) v^2

KE = 5.9192 J

v = 2.8 m/s

b) conserving momentum during inelastic collision,

0.010 ( u) = ( 1.51) (2.8)

u = 422.8 m/s apprx

c) force = mu /t = (0.010 ) ( 422.8) / 5 x 10^-3= 845.6 N

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