A ball of mass 522 g starts at rest and slides down a frictionless track, as sho

Question

A ball of mass 522 g starts at rest and slides down a frictionless track, as shown below. It leaves the track horizontally, striking the ground a distance x = 1.05 m from the end of the track after falling a vertical distance h2 = 1.17 m from the end of the track... Picture of Question is attached.

A  ball of mass 522 g starts at rest and slides down a frictionless track, as shown below. It leaves the track horizontally, striking the ground a distance x =1.05 m from the end of the track after falling a vertical distance h2 = 1.17 m from the end of the track... Picture of Question is attached.

Explanation / Answer

A ball of mass 522 g starts at rest and slides down a frictionless track, as shown below. It leaves the track horizontally, striking the ground a distance x = 1.04 m from the end of the track after falling a vertical distance h2 = 1.15 m from the end of the track.

(b) What is the speed of the ball when it leaves the track?

1.15/(1/2g) = t^2 = 0.237, sq-root = t = 0.4866 secs

1.04/0.4866 = 2.12 m/s (answer b)

(a) At what height above the ground does the ball start to move?

Now this is going to be interesting, because the answer you need could be 2 ways, it could be the easy way or the more advanced correct way.

First of all i'll give the easy way answer of which i think it will be, given the questions you are asking.

1/2mv^2 = mgh
taking "g" = 9.8 m/s^2

KE = 0.5 x 0.522 x 2.12^2 = 1.17 J

mgh = 1.17 J

1.17/(mg) = h = 1.17/(0.522 x 9.8) = 0.228 m + 1.15m = 1.41m (answer 1A)

Now here is the correct answer:

Assuming the ball is solid, inertia = (2/5)mr^2

omega = v/r

using this to solve and cancel mass in the equation.

1.4v^2 = 2gh

simplified = 0.7v^2 = gh

0.7 x 2.12^2 = 3.14

3.14/g = h = 0.32 m + 1.16m = 1.32 (correct answer A)



(c) What is the speed of the ball when it hits the ground?

We know from question b that the horizontal velocity = 1.644 m/s

The time down =
1.15/(1/2g) = t^2 = 0.237, sq-root = t = 0.4866 secs

9.8 x 0.4866 = v = 4.769 m/s

2.12^2 + 4.769^2 = v^2 = 5.2189 m/s

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