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vredina [299]
3 years ago
13

A 2,700-kg truck runs into the rear of a 1,000-kg car that was stationary. The truck and car are locked together after the colli

sion and move with speed 3 m/s. Compute how much kinetic energy was "lost" in this inelastic collision.
Physics
1 answer:
Leto [7]3 years ago
3 0

Answer:

6200 J

Explanation:

Momentum is conserved.

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

The car is initially stationary.  The truck and car stick together after the collision, so they have the same final velocity.  Therefore:

m₁ u₁ = (m₁ + m₂) v

Solving for the truck's initial velocity:

(2700 kg) u = (2700 kg + 1000 kg) (3 m/s)

u = 4.11 m/s

The change in kinetic energy is therefore:

ΔKE = ½ (m₁ + m₂) v² − ½ m₁ u²

ΔKE = ½ (2700 kg + 1000 kg) (3 m/s)² − ½ (2700 kg) (4.11 m/s)²

ΔKE = -6200 J

6200 J of kinetic energy is "lost".

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Answer: D

Reduced impact time will increase the impact force

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If a body of mass M moving with a velocity V collide with another body, the kinetic energy of the body is equal to the work done by the body.

That is, K.E = 1/2mv^2 = F × s

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