Question

A ball is rolled up a ramp with an initial velocity of 5 m/s [U]. 4 seconds later, it is moving down the ramp with a velocity of 3 m/s [D]. What is the acceleration of the ball and at when was the ball stationary?

Answer 1

I'm assuming the ramp is frictionless, so the only force acting on the ball is gravity. Normally we would talk about the component of gravity parallel to the ramp (i.e. mgsinθ, where θ is the angle of the ramp), but here it doesn't matter. Think of the ball, moving in 1 dimension along the ramp, like a car, braking and reversing along a straight road.

This means we can consider mgsinθ more generally as "a," acceleration, for this question.

We have initial velocity (vi), final velocity (vf), and time interval (Δt), so since we are looking for acceleration, we can just use the equation a = Δv/Δt. I will use [U] as the positive direction.

a = Δv/Δt
a = (-3 - 5)/4
a = -2 m/s^2

The acceleration is -2 m/s^2, or 2 m/s [D].

Answer 2

The answer that my teacher gave us is 7.2 m/s^2 and it was stationary at 2.5 seconds but i dont know to get these answer. i got the same answer as you.

Answer 3

That acceleration is way too fast. If it is accelerating at 7.2 m/s^2, in one second, the speed must change by over 7 m/s. This means in 4 seconds, the speed would have to change by 28 m/s, but In the question, the speed changes by only 8 m/s. Unless you left out some information, the acceleration definitely is not 7.2 m/s^2.

As for part two of the question (sorry I didn't answer it before), you have vi = 5 m/s, vf = 0 m/s, a = -2 m/s^2. Now we're looking for Δt (which I'll call t since initial time is 0).

t = Δv/a
t = (0 - 5)/(-2)
t = 2.5 s

The ball is stationary at 2.5 s.

Answer 4

Thanks A Lot!!!!