Question

You are a traffic accident investigator. You have arrived at the scene of an accident. Two trucks of equal mass (3,000 kg each) were involved in a rear-end accident at a stop sign. Here is what you know:
- Truck 1 started from rest at the top of a 22-meter hill and coasted toward the intersection.
- Truck 2 was on a flat stretch of road at the bottom of the hill directly in front of Truck 1.
- At the bottom of the hill, Truck 1 was going 20 m/s; Truck 2 was going 35 m/s.
-There were no skid marks left by Truck 2. The collision occurred at the stop sign where Truck 2 had stopped.
- After the collision, both trucks moved together in the same direction at 10 m/s before slowly rolling to a stop.

1. What was the kinetic energy of each truck before they began braking?

2(a). How much energy did Truck 1 lose from the top to the bottom of the hill?

(b) Where do you suppose that energy went?

Answer 1

Answer:

1) Kinetic energy of the truck 1 is 600,000 joules.

Kinetic energy of the truck 2 is 1,837,500 joules.

2 a)Potential energy of truck 1 is 646,800 Joules.

b) Energy lose by the truck 1 just got converted into heat and sound energy during the collision.

Step-by-step explanation:

Mass of the truck 1 and truck 2,m = 3000 kg

Velocity of the truck 1, $v_1$ = = 20 m./s

Velocity of the truck 2,$v_2$ = 35 m./s

1. Kinetic energy of the truck 1:$\frac{1}{2}mv_{1}^2=\frac{1}{2}\times 3000 kg\times (20 m/s)^=600,000 joules$

Kinetic energy of the truck 2:$\frac{1}{2}mv_{2}^2=\frac{1}{2}\times 3000 kg\times (35 m/s)^=1,837,500 joules$

2.a) Energy loose by the truck 1:

Initial height of the truck on the hill = 22 m

At that height potential energy will be maximum:

Potential energy of truck 1 at that height = $mgh=3000\times 9.8 m/s^2\times 22 m=646,800 Joules$

b) Energy lose by the truck 1 just got converted into heat and sound energy during the collision.