Question

A 1250 kg car is moving down the highway with a velocity of 32.0 m/s when it bumps into the car ahead of it which has a mass of 875 kg and a velocity of 25.0 m/s. After the collision, the two cars stick together. What will be the resulting velocity of the two cars together? How much energy will be lost in this collision?

Answer 1

A. If the collision is perfectly inelastic then it follows the equation,
    
             m1v1 + m2v2 = (m1 + m2)(v3)

Substituting,

           (1250 kg)(32 m/s) + (875 kg)(25 m/s) = (1250 kg + 875 kg)(v3)

The value of v3 from the equation is 29.12 m/s. 

B. The kinetic energy is calculated through the equation,

              KE = 0.5mv²

Using this equation to solve for the total kinetic energies before and after the collision,

    Before collision:

         KE = 0.5(1250 kg)(32 m/s)² + (0.5)(875 kg)(25 m/s)²
             KE = 913437.5 J

    After collision:
           KE = (0.5)(1250 kg + 875 kg)(29.12 m/s)²
                 KE = 900972.8 J

The difference is equal to 12464.7 J

Answer 2

Resulting velocity = 29.1 m/s

Lost energy = 13701.875 j ≈ 13702 j

The collision is elastic so, the final velocity of the car will be the same. Therefore,

v = m₁v₁ + m₂v₂/m₁ + m₂

v = final velocity

m₁ = mass of car1 = 1250 kg

m₂ = mass of car2 = 875 kg

v₁ = initial velocity of car1  = 32 m/s

v₂ = initial velocity of car2 = 25 m/s

v = (1250×32) + (875×25)/ 1250 + 875

v = 40000 + 21875/2125

v = 61875 / 2125

v = 29.1176470588 ≈ 29.1m/s

Let's calculate the lost kinetic energy.

The before kinetic energy will  be

KE₁ + KE₂ = total KE

KE₁  =1/2×1250×32²= 640000 j

KE₂ = 1/2 × 875 × 25²  = 273437.5  j

total = 640000 j + 273437.5  j = 913437.5 j

After kinetic energy

1/2×(1250+875)×29.1² = 899735.625 j

Lost energy  = 913437.5 j -  899735.625 j = 13701.875 j ≈ 13702 j

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