Question

Kelly is driving a car with a mass of 800 kilograms in a south direction with a speed of 18 meters/second. Dominic is driving a truck with a mass of 1500 kilograms with a velocity of 15 meters/second to the east. They undergo perfectly inelastic collision. What is the magnitude of the final momentum of both the truck and the car.

Answer 1

Approximately
`2.7* 10^(4)\; \rm kg \cdot m \cdot s^(-1)`.

Step-by-step explanation:

The two colliding objects would be stuck to each other if the collision is perfectly inelastic. While kinetic energy is conserved only in elastic collisions, momentum is conserved in all types of collisions. Therefore, the momentum of the truck and the car right after the collision will be equal to the vector sum of their momentum right before the collision.

Momentum of the car right before the collision:

`\vec{p}(\text{car, before}) = m \cdot \vec{v} = 800\; \rm kg * 18\; \rm m \cdot s^(-1)= 14400\; \rm kg \cdot m \cdot s^(-1)`.

Momentum of the truck right before the collision:

`\vec{p}(\text{truck, before}) = m \cdot \vec{v} = 1500\; \rm kg * 15 \; \rm m \cdot s^(-1) = 22500\; \rm kg \cdot m \cdot s^(-1)`.

Note that these two momentum are perpendicular to each other. The magnitude of their vector sum should be found using the Pythagorean theorem:

`\begin{aligned}&\left|\vec{p}(\text{sum, before})\right| \\ &= \left|\vec{p}(\text{car, before}) + \vec{p}(\text{truck, before})\right| \\ &= \left(√(14400^2 + 22500^2)\right)\; \rm kg \cdot m \cdot s^(-1) \approx 2.7* 10^(4)\; \rm kg \cdot m \cdot s^(-1) \end{aligned}`.

Because momentum is conserved, the sum of the momentum of the car and the truck right after the collision should be equal to the sum right before the collision. In other words:

`\vec{p}(\text{sum, after}) = \vec{p}(\text{sum, before})`.

Therefore:

`\left|\vec{p}(\text{sum, after})\right| = \left|\vec{p}(\text{sum, before})\right| \approx 2.7 * 10^(4)\; \rm kg \cdot m \cdot s^(-1)`.