Question

A 75-g projectile traveling at 600 m/s strikes and becomes embedded in a 50-kg block, which is initially stationary. Compute the kinetic energy lost during the impact. Express your answer as an absolute value |AE| and as a percentage n of the original system energy E.

Answer 1

Answer:

Change in kinetic energy: 13479.77 J

Percentage change in kinetic energy = 99.85%

Explanation:

Change in momentum when the projectile becomes embedded in the block will be 0.

This means:

$M_P *V_P + M_B * V_B = (M_p+M_B)*V_P_+_B$

here subscript P means projectile, and subscript B means block.

Solving this equation we get:

(0.075 * 600) + (50 * 0) = (0.075 + 50) * V

$V_P_+_B = 0.899 m/s$

Using this velocity we can compute the change in kinetic energy:

Initial kinetic energy = 0.5 * m * v^2

Initial kinetic energy = 0.5 * 0.075 * 600^2

Initial kinetic energy = 13500 J

Final Kinetic energy = 0.5 * m * v^2

Mass (m) = mass of block + mass of projectile = 50.075 kg

Final velocity v = 0.899 m/s

Final Kinetic Energy = 0.5 * 50.075 *  0.899^2

Final Kinetic Energy = 20.235 J

Change in kinetic energy = 13500 - 20.235 = 13479.77 J

Percentage change of kinetic energy = (13479.77 / 13500) * 100

Percentage change of kinetic energy = 99.85 %