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 = 0.93 m from the end of the track after falling a vertical distance h2 = 1.18 m from the end of the track.

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 = 0.93 m from the end of the track after falling a vertical distance h2 = 1.18 m from the end of the track.

(a) At what height above the ground does the ball start to move?
m
(b) What is the speed of the ball when it leaves the track?
m/s
(c) What is the speed of the ball when it hits the ground?
m/s

Answer 1

To find the answers to these questions, we can use the principles of conservation of energy and motion.

(a) To find the height above the ground where the ball starts to move, we need to equate the potential energy at that height with the kinetic energy when the ball starts moving. We can use the formula:

mgh1 = (1/2)mv^2

Here, m is the mass of the ball, g is the acceleration due to gravity, h1 is the height above the ground where the ball starts to move, and v is the speed of the ball. Since the ball starts at rest, its initial speed is zero. Rearranging the formula, we have:

h1 = (1/2)v^2 / g

(b) To find the speed of the ball when it leaves the track, we can use the principle of conservation of energy. The potential energy at the starting height (h1) gets converted into kinetic energy at the end of the track. So, we can use the formula:

mgh1 = (1/2)mv^2

Here, m, g, h1, and v are the same as in the previous equation. Rearranging the formula, we have:

v = sqrt(2gh1)

(c) To find the speed of the ball when it hits the ground, we need to calculate the kinetic energy of the ball just before it hits the ground. We can use the formula:

mgh2 = (1/2)mv^2

Here, m, g, h2, and v are the same as in the previous equations. Rearranging the formula, we have:

v = sqrt(2gh2)

Now let's substitute the given values into the formulas to find the answers.

Given values:
m = 522 g = 0.522 kg
x = 0.93 m
h2 = 1.18 m
g = 9.8 m/s^2

(a) To find the height above the ground where the ball starts to move (h1):
h1 = (1/2)(0.93)^2 / 9.8 = 0.044 m

(b) To find the speed of the ball when it leaves the track (v):
v = sqrt(2 * 9.8 * 0.044) = 0.94 m/s

(c) To find the speed of the ball when it hits the ground (v):
v = sqrt(2 * 9.8 * 1.18) = 5.18 m/s

So, the answers are:
(a) The height above the ground where the ball starts to move is 0.044 m.
(b) The speed of the ball when it leaves the track is 0.94 m/s.
(c) The speed of the ball when it hits the ground is 5.18 m/s.