- #1

- 4

- 0

here is the question......

Q2: The rollercoaster car from Task 1 Q2 above is initially travelling at 6.5m/s at point A in the diagram below. Assuming negligible frictional forces as previously, what will be the final velocity of the car at point B?

The image shows that the cart drops by 3.5 m on a ramp, and point b is further along the track.

So here is what I did:

At the top of the drop the drop the carriage already has an amount of kinetic energy, and potential energy, the potential energy will add on to the kinetic energy as it begins to accelerate down the drop.

Initial Kinetic Energy = 0.5 x mass x velocity^2 = 0.5x440x6.5^2 = 9295 joules

Potential energy = mass x gravity x height = 440x9.8x3.5 = 15092 joules

Total Energy at bottom of drop = 15092+9295= 24387 joules.

Transpose this formula to find V: KE= 0.5 x mass x v^2

V = √2xKinetic Energy/Mass

V = √2x24387/440

V = 10.53 m/s

Now here is my question why is it that I have to calculate the energies to find the velocity rather than calculate the velocities individually, i.e. find the velocity of the carriage dropping using the initial velocity as 0.