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 ?

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1 Answer
Jan 19, 2016

(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