An 8 g bullet from an AK-47 machine gun is shot at a speed of 750 m/s towards a

Question

An 8 g bullet from an AK-47 machine gun is shot at a speed of 750 m/s towards a 2.4 kg block of wood which is at rest on a horizontal frictionless surface. The bullet exits the block of wood with half its initial speed, or 375 m/s.
(a) What is the speed of the block of wood after the bullet exits?



m/s

(b) What is the kinetic energy of the block of wood after the collision?
J

(c) The block of wood slides along the frictionless surface until it hits a rough section which has a kinetic coefficient of friction of 0.2. How far along the rough surface does the block of wood slide before it comes to a stop?
m

Explanation / Answer

a) conserving energy,
(1/2)mv1^2 = (1/2)m*v2^2 + (1/2)m2*v^2
> 0.008*750^2 = 0.008*375^2 + 3*v^2
=> v = 33.54 m/s

b) KE = (1/2)3*v^2 = 1687.5 J

c) v^2 - u^2 = -2as

=> 33.54^2 = 2*0.2*9.81*s

=> s = 286.7 m

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