a 6.25-gram bullet traveling at 365 ms strikes and enters a 4.50-kg crate. The crate slides 0.15 m along a wood floor until it comes to rest. What is the change in kinetic energy of the system after the collision

the change in kinetic energy of the system is 0.577 J

Explanation:

Given;

mass of the bullet, m₁ = 6.25 g = 0.00625 kg

initial velocity of the bullet, u₁ = 365 m/s

mass of the crate, m₂ = 4.5 kg

initial velocity of the crate, u₂ = 0

distance moved by the system after collision, d = 0.15 m

Determine the final velocity of the system after collision;

m₁u₁ + m₂u₂ = v (m₁ + m₂)

0.00625 x 365 + 4.5 x 0 = v(0.00625 + 4.5)

2.2813 + 0 = v(4.5063)

2.2813 =  v(4.5063)

v = 2.2813 / 4.5063

v = 0.506 m/s

The change in kinetic energy of the system after collision is calculated as;

ΔK.E = ¹/₂ (m₁ + m₂)v²

ΔK.E =  ¹/₂ (4.506) x 0.506²

ΔK.E = 0.577 J

Therefore, the change in kinetic energy of the system is 0.577 J

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