Question

2200 kg semi truck driving down the highway has lost control. The truck rolls across the median and into oncoming traffic. The truck collides with a 2000 kg car that is traveling north at 45 m/s. The two vehicles stick together creating a single wreck that travels south at 15 m/s. what was the initial speed and direction of the semi truck?

Answer 1

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Step-by-step explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

`m_(S)\cdot v_(S)+m_(C)\cdot v_(C) = (m_(S)+m_(C))\cdot v` (1)

Where:

`m_(S)`,
`m_(C)` - Masses of the semi truck and the car, measured in kilograms.

`v_(S)`,
`v_(C)` - Initial velocities of the semi truck and the car, measured in meters per second.

`v` - Final speed of the system after collision, measured in meters per second.

If we know that
`m_(S) = 2200\,kg`,
`m_(C) = 2000\,kg`,
`v_(C) = 45\,(m)/(s)` and
`v = -15\,(m)/(s)`, then the initial velocity of the semi truck is:

`m_(S)\cdot v_(S) = (m_(S)+m_(C))\cdot v -m_(C)\cdot v_(C)`

`v_(S) = ((m_(S)+m_(C))\cdot v - m_(C)\cdot v_(C))/(m_(S))`

`v_(S) = \left(1+(m_(C))/(m_(S)) \right)\cdot v - (m_(C))/(m_(S)) \cdot v_(C)`

`v_(S) = v +(m_(C))/(m_(S))\cdot (v-v_(C))`

`v_(S) = -15\,(m)/(s)+\left((2000\,kg)/(2200\,kg) \right) \cdot \left(-15\,(m)/(s)-45\,(m)/(s) \right)`

`v_(S) = -69.545\,(m)/(s)`

The semi truck travels at an initial speed of 69.545 meters per second downwards.