In a physics experiment, Willy rolls a metal ball down a sloped track from a height of 25cm. At the bottom end

of the track, it slopes upwards at an angle of 53 degrees. He finds that if he gives the ball a push, he can make it rise above the height of the top end of the ramp. Explain.

What am I supposed to prove?

In a physics experiment, Willy rolls a metal ball down a sloped track from a height of 25cm. At the bottom end of the track, it slopes upwards at an angle of 53 degrees. He finds that if he gives the ball a push, he can make it rise above the height of the top end of the ramp. Explain.


This problem deals with conservation of energy.
As the ball moves down the track, its kinetic energy increases as the potential energy decreases. As the ball moves upward, its potential energy increases as the kinetic energy decreases.
The maximum height of the car after moving down and up, can’t be greater than the initial height of the car, unless the car get’s some extra energy. If he gives the ball a push, the kinetic energy of the ball will increase. So the car may have enough energy to rise above the height of the top end of the ramp.

Example:
PE = m * g * h
KE = ½ * m * v^2

Mass of ball = 500g = 0.5 kg
Initial height = 25 cm = 0.25 m
Initial PE = 0.5 * 9.8 * 0.25 = 1.225 J
At the bottom end of the track, the ball’s KE = 1.225 J
1.225 = ½ * 0.5 * v^2
Velocity at bottom of track = (1.225 ÷ 0.25)^.5 = 2.21 m/s

As the ball rolls up the 53˚ incline, the velocity decreases, due to the force parallel = m* g * sin 53˚
Acceleration = Force parallel ÷ m = -9.8 * sin 53˚ = -12.3 m/s^2
Time to stop = (Final velocity – Initial velocity) ÷ a = (0 – 2.21) ÷ -12.3 = 0.18 seconds.

Now let’s push the ball!
A force of 100 N is applied to the car for 0.01 meters as it begins to move down the 0.25 meters

Work = force * distance = 100 * 0.01 = 1 N*m = 1 Joule of energy
Now the car has total energy of 1 + 1.225 = 2.225 J when it reaches the bottom of the track.

Velocity at bottom of track = (2.225 ÷ 0.25)^.5 = 2.98 m/s
Acceleration = -12.3 m/s^2
Time to stop = (0 – 2.98) ÷ -12.3 = 0.24 seconds

The energy from the “push” allowed the ball to move up incline for an extra 0.6 seconds. So the ball moves farther up the ramp.
SO
The ball maximum height after moving down to the bottom and up the incline is greater than the initial height!

You are supposed to prove that if the ball begins rolling with some amount of velocity it will reach a maximum height of greater than 25 cm! You can prove this by finding the force of his little "push" required o achieve this