Question

In the figure, a frictionless roller coaster car of mass m = 805 kg tops the first hill with speed vi = 18 m/s at height h = 41 m. What is the speed of the car at (a) point A, (b) point B, and (c) point C? (d) How high will the car go on the last hill ?

Answer 1

(a) At point A `=18 m//s`
(b) At point B `approx 26.94 m//s`
(c) At Point C `approx 33.58m//s`
(d) `41 m`

Explanation:

The problem is about Conservation of Energy, between Potential Energy and Kinetic energy changes.
Given is friction-less roller coaster. it is also assumed that there is no loss of energy due to air friction.

Given also is that initial velocity is towards the `x` axis only.
Therefore, at any location `x`, speed `=sqrt (v_x^2+v_y^2)`
`=sqrt (18^2+v_y^2)`
Inspection reveals that `y` component of velocity is created due to change of `PE=mgh`, in to `KE=1/2 m v^2`. At any point `x`
Initial `PE=mgh=PE(x)+KE(x)`

Considering only the `y` component of velocity of the car.

(a) At point A. The situation is same as at `t=0`, where `v_y =0`.

(b) At point B. Height is `h//2,:.`half of initial `PE` has got converted into `KE` of the car
`mgh/2=KE_`(B)`=1/2 m v_B^2`
`implies v_B^2=gh`
`implies v_B^2=9.8 xx 41=401.8`, for y direction only
`V_(Total)`(B)`=sqrt(324+401.8)`

(c) At Point C. Here all the initial `PE` has got converted into `KE` of the car.
Following similar steps as in for point B. above,

`V_(Total)`(C)`=sqrt(324+803.6)`

(d) At the last hill. Car can go only up to initial height `h`