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Question
A bullet of mass 15 g has a speed of 400 m/s. What is its kinetic energy? If the bullet strikes a thick target and is brought to rest in 2 cm, calculate the average net force acting on the bullet. What happens to the kinetic energy originally in the bullet?
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Solution
Mass of bullet, (m) = 0.015 kg
Initial velocity of bullet, (v1) = 400 m/s
Distance, (S) = 0.02 m
So, initial kinetic energy of bullet,
`K.E = 1/2 mv^2`
Therefore, initial kinetic energy,
`(K.E)_1 = 1/2(0.015)(400)^2` J
= 1200 J
Similarly, final kinetic energy as the final velocity is zero,
(K.E)2 = 0
So,
Work done = Change in kinetic energy
Therefore work done,
Work done = (K.E)2 – (K.E)1
(Force) (Distance) = (0 –1200) J
So,
`"Force" = -1200/0.02`
= -60000
= `-(6)(10)^4` J
Negative sign shows that the force applied is opposite to the direction of motion of the bullet.
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