A thin block of soft wood with a mass of 0.082 kg rests on a horizontal friction

Question

A thin block of soft wood with a mass of 0.082 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 641 m/s at a block of wood and passes completely through it. The speed of the block is 21 m/s immediately after the bullet exits the block.

(a) Determine the speed of the bullet as it exits the block.

(b) Determine if the final kinetic energy of this system (block and bullet) is equal to, less than, or greater than the initial kinetic energy.

(c) Verify your answer to part (b) by calculating the initial and final kinetic energies of the system.

KEi =

KEf =

Explanation / Answer

a)

m = mass of bullet = 0.00467 kg

vbi = velocity of bullet before collision = 641 m/s

vbf = velocity of bullet after collision = ?

M = mass of block = 0.082 kg

vBi = velocity of Block before collision = 0 m/s

vBf = velocity of Block after collision = 21 m/s

using conservation of momentum

m vbi + M vBi = m vbf + M vBf

(0.00467) (641) + (0.082) (0) = (0.00467) vbf + (0.082) (21)

vbf = 272.3 m/s

b)

KEi = initial KE = (0.5) (m v2bi + M v2Bi ) = (0.5) ((0.00467) (641)2 + (0.082) (0)2) = 959.41 J

KEf = Final KE = (0.5) (m v2bi + M v2Bf ) = (0.5) ((0.00467) (272.3)2 + (0.082) (21)2) = 191.21 J

hence final KE < initial KE

c)

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