Consider the air track in Figure 14. Suppose a gliding cart with a mass of 0.5 kg bumps into, and sticks to, a stationary cart that has a mass of 1.5 kg. If the speed of the gliding cart before impact is v_{before} , how fast will the coupled carts glide after collision?
Question 5.S&C.5: Consider the air track in Figure 14. Suppose a gliding cart ...
According to momentum conservation, the momentum of the 0.5-kg cart before the collision = momentum of both carts stuck together afterward.
\begin{aligned}&0.5 kg v_{\text {before }}=(0.5 kg +1.5kg ) v_{\text {after }} \\&v_{\text {after }}=\frac{0.5 kg v_{\text {before }}}{(0.5 kg +1.5 kg )}= \\&\quad \frac{0.5 v_{\text {before }}}{2}=\frac{v_{\text {before }}}{4} \text {[note kg in equation] }\end{aligned}This makes sense, because four times as much mass will be moving after the collision, so the coupled carts will glide more slowly. The same momentum means four times the mass glides 1/4 as fast. So we see that changes in an object’s motion depend both on force and on how long the force acts. When “how long” means time, we refer to the quantity “force \times time” as impulse. But “how long” can mean distance also. When we consider the quantity “force\times distance,” we are talking about something entirely different—the concept of energy.
Related Answered Questions
1. a. Neglecting inefficiencies, the entire initia...
1. a. Its momentum hitting the ground is its [late...
From the definition that work = force \time...
1. Nine times farther. The car has nine times as m...
1. Yes, because no acceleration means that no chan...
1. In accordance with Newton’s third law, the forc...
1. The force of impact will be only a third of wha...
1. Both have the same momentum (1 \text { t...