A 20-g bullet moving at 1000 m/s is fired through a 1kg block of wood emerging at a speed of 100m/s.what is the change in kinetic energy of the bullet

Answer:
D. 0.16 KJ
Explanation:
Given the following data;
Initial velocity, U = 1000 m/s
Mass of bullet, M1 = 20 g to kilograms = 20/1000 = 0.02 kg
Mass of block, M2 = 1 kg
Final velocity, V = 100 m/s
To find the change in kinetic energy;
First of all, we would determine the velocity after the collision by applying the law of conservation of momentum.
[tex] M_{1}U - M_{1}V = M_{2}V_{f} [/tex]
Substituting into the above formula, we have;
[tex] 0.02*1000 - 0.02*100 = 1*V_{f} [/tex]
[tex] 20 - 2 = V_{f} [/tex]
Vf = 18 m/s
Next, we would find the kinetic energy of block-bullet;
[tex] K.E = \frac{1}{2}MV^{2}[/tex]
Substituting into the formula, we have;
[tex] K.E = \frac{1}{2}*1*18^{2}[/tex]
[tex] K.E = \frac{1}{2}*324 [/tex]
K.E = 162 Joules.
In Kilojoules;
K.E = 162/1000
K.E = 0.162 ≈ 0.16 KJ