A paving truck with a mass of 5.0 times 10 to the third kilograms drives in a straight line at 2.5 meters per second and dumps 1.0 times 10 to the third kilograms of gravel on the road, taking negligible time. What is the speed of the truck after dumping the gravel?
We were told that the mass of the truck with the gravel together is equal to 5.0 times 10 to the third kilograms. We’ll call that 𝑚 sub 𝑖. The truck is moving at a speed of 2.5 meters per second. We’ll call that 𝑣 sub 𝑖. While moving at this speed, the truck dumps 1.0 times 10 to the third kilograms of gravel on the road. We’ll call that 𝑚 sub 𝑔. We want to know the speed of the truck after it dumbs the gravel. We’ll call that 𝑣 sub 𝑓.
Now just thinking of this problem in terms of masses and speeds, we might expect that if the engine of the truck is working at the same power level throughout, that less mass will mean more speed. To approach this problem, we’ll rely on the conservation of momentum. The conservation of momentum simply states that for any interaction, the initial momentum of the objects involved is equal to the final momentum of the objects. And we can recall further that momentum 𝑝 is defined as an object’s mass times its velocity.
So for our case with the dump truck, we can write that the initial momentum of the system is equal to the final momentum; 𝑝 sub 𝑖 is equal to 𝑝 sub 𝑓. Expanding that out, 𝑚 sub 𝑖, the initial mass, times 𝑣 sub 𝑖, the initial speed, is equal to the product of the final mass, 𝑚 sub 𝑓, times 𝑣 sub 𝑓, the final speed. It’s 𝑣 sub 𝑓 we want to solve for. So to get there, let’s divide both sides of our equation by 𝑚 sub 𝑓, the final mass of the truck. Doing so, cancels that term out of the right side of our equation. And we see that 𝑣 sub 𝑓 is equal to the ratio of the initial to the final mass of the track multiplied by the initial speed of the truck. We’re given 𝑚 sub 𝑖, the truck’s initial mass. We can solve for 𝑚 sub 𝑓 because that’s equal to the initial mass of the truck, 𝑚 sub 𝑖, minus the mass of the gravel it dumbs. 5.0 times 10 to the third kilograms minus 1.0 times 10 to the third kilograms equals 4.0 times 10 to the third kilograms. That’s the final mass of the truck after it dumbs the gravel.
We can now plug in for 𝑚 sub 𝑖, 𝑚 sub 𝑓, and 𝑣 sub 𝑖. With all these values inserted, when we multiply and divide to solve for 𝑣 sub 𝑓, we find that it’s a speed of 3.1 meters per second. This is the final speed of the truck after it dumbs the gravel. So we see that our intuition was correct. A lighter mass under the same power has more speed. This is the conservation of momentum.