Q#21 (Past Exam Paper – November 2016 Paper 11 & 13 Q6)
A cyclist pedals along a raised horizontal track. At the end of the track, he travels horizontally into the air and onto a track that is vertically 2.0 m lower.
The cyclist travels a horizontal distance of 6.0 m in the air. Air resistance is negligible.
What is the horizontal velocity v of the cyclist at the end of the higher track?
A 6.3 m s-1 B 9.4 m s-1 C 9.9 m s-1 D 15 m s-1
Solution:
Answer: B.
This is a projectile motion question. So, the horizontal and vertical component of the motion can be considered separately. The quantity that can relate both motion is the time t.
Initially, since the cyclist is on a horizontal track, he has only a horizontal velocity of v, and no vertical velocity.
Air resistance is negligible. So, the only force (causing an acceleration) that affects the motion is the force of gravity (which is downwards).
Consider the vertical component:
Initial velocity, u = 0
Acceleration, a = 9.81 m s-2 (due to gravity)
Distance, s = 2.0 m
Time, t = ??? (time to reach the lower horizontal track)
s = ut + ½ at2
2 = 0 + ½ × 9.81 × t2
Time t = √(2 × 2 / 9.81) = 0.64 s
So, the cyclist takes 0.64 s to fall a vertical distance of 2.0 m. During this 0.64 s, he also moves a horizontal distance of 6.0 m simultaneously.
Consider the horizontal component:
Acceleration, a = 0
Initial speed = final speed = v
Distance travelled = 6.0 m
Time taken = 0.64 s
Speed v = distance / time = 6.0 / 0.64
Sped v = 9.4 m s-1
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