Suppose you build a straight frictionless track from the origin to the point (x,

Question

Suppose you build a straight frictionless track from the origin to the point (x,y) = (13.79 m, -1.93 m). Define your coordinate system so that gravity points in the negative y direction. A particle is released from rest at the origin. What is the speed of the particle when it reaches the end of the track?

How much time does it take for the particle to reach the end of the track?

Now suppose you change the shape of the track so that the particle will travel between the two points in the shortest amount of time possible. What is the speed of the particle when it reaches the end of the track now?

How much time does it take for the particle to reach the end of the track now?

Explanation / Answer

from origin to (13.79,-1.93) m, a fricitonless track is built, and g acts along -y axis

a. when the particel is released at rest from the origin

Loss in PE when particle reaches end of track = mgh

h = 1.93 m

so Loss of PE = mg*1.93

from work energy theorem

gain in KE = loss in PE

0.5mv^2 = mg*1.93 ( where v is final speed of particel at the end of the track)

v^2 = 2g*1.93

v = 6.153 m/s

b. time taken to reach the end of the track = t

acceleration along the track, a = gsin(theta)

where theta = arctan(1.93/13.79) = 7.967 deg

so a = 1.359 m/s/s

length of the path = sqroot(13.79^2 + 1.93^2) = 13.9244 m

so time taken = t

13.9244 = 0.5*1.359*t^2

t = 4.526 s

c. speed of the particle will not depend on the shape of the track if it is frictionless, as PE depends only on end points and not on the path

so v = 6.153 m/s

d. vertical distance moved, y = -1.93 m

so in time t, vertical speed, vy = ?

2*(-1.93)*(-9.81) = vy^2

vy = 4.351 m/s

so, horizontasl speed at the end of journey = vx

6.153^2 = vx^2 + vy^2 = vx^2 + 4.351^2

vx = 4.350 m/s

horizontal distance, x = 13.79 m

time taken, t = x/vx = 3.169 s

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