Question

A 1 225.0 kg car traveling initially with a speed of 25.000 m/s in an easterly direction crashes into the back of a 9 700.0 kg truck moving in the same direction at 20.000 m/s. The velocity of the car right after the collision is 18.000 m/s to the east.(a) What is the velocity of the truck right after the collision? (Give your answer to five significant figures.) m/s east (b) What is the change in mechanical energy of the cartruck system in the collision? J (c) Account for this change in mechanical energy.

Answer 1

Answer:

The answers to the questions are;

(a) The velocity of the truck right after the collision is 20.884 m/s

(b) The change in mechanical energy of the car truck system in the collision is -9076.4384 J

(c) The change in mechanical energy is due to energy consumed by the collision process.

Explanation:

(a) From the principle of conservation of linear momentum, we have

m₁·v₁+m₂·v₂ = m₁·v₃ + m₂·v₄

Where:

m₁ = Mass of the car = 1225.0 kg

m₂ = Mass of the truck = 9700.0 kg

v₁ = Initial velocity of the car = 25.000 m/s

v₂ = Initial velocity of the truck = 20.000 m/s

v₃ = Final velocity of the car right after collision = 18.000 m/s

v₄ = Final velocity of the truck right after collision

Therefore

1225.0 kg × 25.000 m/s  +  9700.0 kg × 20.000 m/s = 1225.0 kg × 18.000 m/s  + 9700.0 kg × v₄

That is 30625 kg·m/s + 194000 kg·m/s = 22050 kg·m/s + 9700.0 kg × v₄

Making v₄ the subject of the formula yields

v₄ = (202575 kg·m/s)÷9700.0 kg = 20.884 m/s

The velocity of the truck right after the collision to five significant figures = 20.884 m/s

(b) The change in mechanical energy of the car truck system in the collision can be found by

The change in kinetic energy of the car truck system

Change in kinetic energy, ΔK.E. = Sum of final kinetic energy - Sum of initial kinetic energy

That is ΔK.E. = ∑ Final K.E -∑ Initial K.E.

ΔK.E. = $(\frac{1}{2} m_1v_3^{2}+\frac{1}{2} m_2v_4^{2}) - (\frac{1}{2} m_1v_1^{2} +\frac{1}{2} m_2v_2^{2} )$

         = ($\frac{1}{2}$·1225·18²+ $\frac{1}{2}$·9700·20.884²) - ($\frac{1}{2}$·1225·25²+$\frac{1}{2}$·9700·20²)

         = 2313736.0616 kg·m²/s² - 2322812.5 kg·m²/s² =  -9076.4384 kg·m²/s²

1 kg·m²/s² = 1 J ∴ -9076.4384 kg·m²/s² = -9076.4384 J

(c) The energy given off by way of the 9076.4384 J is energy transformed into other forms including

1) Frictional resistance between the tires and the road for the truck and car

2) Frictional resistance in the transmission system of the truck to increase its velocity

3) Sound energy, loud sound heard during the collision

4) Energy absorbed when the car and the truck outer frames are crushed

5) Heat energy in the form of raised temperatures at the collision points of the car and the truck.

6) Energy required to change the velocity of the car over a short distance.