Question 21
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
Reference: Past Exam Paper – November 2016 Paper 11 & 13 Q6
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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