A 60-g projectile traveling at 605 m/s strikes and becomes embedded in the 54-kg block, which is initially stationary. Compute the energy lost during the impact. Express your answer as an absolute value |ΔE| and as a percentage

Kinetic energy lost in collision is 1097.95 J

Explanation:

Given,

Mass,  = 60 g = 0.006 kg

Speed, = 605 m/s

= 54 kg

= 0

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

So,

Before collision, the kinetic energy is

Therefore, kinetic energy before collision is 1098 J

Kinetic energy after collision:

By plugging in the values, we get

Since,

initial Kinetic energy = Final kinetic energy

1098.075 J = 0.1212 J + K×E(lost)

K×E(lost) = 1098.075 J - 0.121 J

K×E(lost) = 1097.95 J

Therefore, kinetic energy lost in collision is 1097.95 J

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