Physics Question

[Caffine]

A ball is initially rolling up a slight incline at 0.2 m/s. It decelerates uniformly at 0.05 m/s^2.

1.) At what time does the ball come to a stop?
2.) What is the ball's net displacement after 6s?

---

Members don't see this ad.

 

[Charles_Carmichael]

Why don't you tell us how you tried to solve the questions first? Then, we can help you figure out where you went wrong.

 

[Caffine]

Why don't you tell us how you tried to solve the questions first? Then, we can help you figure out where you went wrong.

The reason why I didn't explain how I solved this problem was because I wanted to see how others interpreted the question. The answer I obtained was different from the solution. For instance, the first question asks - "what time does the ball come to a stop." I figured, well since it's rolling up an incline, it'll reach 0 m/s at it's peak height, but it's still accelerating downwards. The ball actually stops when it rolls back down. It takes 4 seconds to reach peak height, and 4 seconds to reach the bottom ..so I figured the total time would be 8 seconds. They wen't with 4 seconds instead.

For the second question, they plugged in numbers into one of the kinematics equations:

displacement = vt + 1/2at^2
displacement = 0.2(6) - 1/2(0.5)(6)
displacement = 0.3 meters

But this would only make sense if they were asking for "total distance" not "net displacement" if you consider what I said before, the net "displacement" would then be:

displacement = 1/2(0.5)(2)^2
displacement = 0.1 m (this wasn't an option)

The reason I put 2 seconds instead of 6 was because after 4 seconds (when the object reaches peak height), the ball begins traveling down. t = 6 seconds is the same thing as t = 2 seconds as it's traveling downwards.

 

[Rabolisk]

The ball actually stops when it rolls back down. It takes 4 seconds to reach peak height, and 4 seconds to reach the bottom ..so I figured the total time would be 8 seconds. They wen't with 4 seconds instead.

You don't know that the ball ever stops once it starts rolling downhill, not considering friction. There is no reason to believe that the incline is pushed up against a wall which necessarily stops the ball. The only thing you know is that when velocity = 0, by definition, the ball has stopped, even momentarily.

But this would only make sense if they were asking for "total distance" not "net displacement"

No, it works for net displacement, not total distance. The kinematics equations explicitly takes into account that you are talking about displacement in one dimension (i.e. on a line). The net displacement is 0.3 meters, and the total distance traveled is 0.5 meters. The ball travels 0.4 meters up the hill, then 0.1 meters down the hill to give net displacement of 0.3 meters and total distance of 0.5 meters. If you did two independent equations with t = 4, v0 = 0.2 (1) and t = 2, v0 = 0, then you will see that this is true.

 

[Caffine]

You don't know that the ball ever stops once it starts rolling downhill, not considering friction. There is no reason to believe that the incline is pushed up against a wall which necessarily stops the ball. The only thing you know is that when velocity = 0, by definition, the ball has stopped, even momentarily.



No, it works for net displacement, not total distance. The kinematics equations explicitly takes into account that you are talking about displacement in one dimension (i.e. on a line). The net displacement is 0.3 meters, and the total distance traveled is 0.5 meters. The ball travels 0.4 meters up the hill, then 0.1 meters down the hill to give net displacement of 0.3 meters and total distance of 0.5 meters. If you did two independent equations with t = 4, v0 = 0.2 (1) and t = 2, v0 = 0, then you will see that this is true.

Thanks for clarifying. Makes sense!