Question

Regarding the Law of Conservation of Momentum, suppose a hand tool is dropped into some freshly poured cement. The tool has a mass of 1.1 kg. The velocity at the moment of impact (v1) is 2 meters per second. At that moment, the momentum may be computed as M1 = mv1 M1 = 1.1 kg x 2 m/s = 2.2 kg m/s. The dropped tool sinks a tiny bit into the cement and stops in 0.05 s, at which time the momentum is zero. Include in answers the units. What is the change in the momentum of the object?

Answer 1

The change in momentum of the object is -2.2 kg·m/s, signifying that the tool came to a complete stop from its initial motion.

Step-by-step explanation:

To determine the change in momentum of the object, we look at the initial and final momentum. At the moment of impact, the tool has a momentum (M1) of 2.2 kg·m/s. When it stops, the final momentum (M2) is 0 kg·m/s, because the velocity of the tool is zero. Therefore, the change in momentum (\(ΔM\)) is the final momentum minus the initial momentum.

\(ΔM = M2 - M1\)

\(ΔM = 0 kg·m/s - 2.2 kg·m/s\)

\(ΔM = -2.2 kg·m/s\)

The change in momentum of the object is therefore -2.2 kg·m/s. The negative sign indicates that the change in velocity was in the opposite direction of the initial velocity, which means the tool came to a stop.