Question 20
A ball is released
from rest above a horizontal surface. It strikes the surface and bounces
several times.
The velocity-time graph
for the first two bounces is shown.
What is the maximum
height of the ball after the first bounce?
A 0.20 m B 0.25 m C 0.45 m D 0.65 m
Reference: Past Exam Paper – November 2015 Paper 12 Q9
Solution:
Answer: A.
Initially, the ball is at
rest above the horizontal surface. In the graph, this is the point (0, 0).
As the ball is released, it
falls down and its velocity increases until its velocity rises to + 3.00 m s-1
in the graph. In the graph, this motion is shown as positive. So, the downward
direction is taken as positive. When the velocity is positive, the ball is
moving downwards.
As the ball strikes the
ball, its velocity decreases instantly to zero (at time = 0.30 s). The ball
then bounces and starts to move upwards with a velocity of – 2.00 m s-1.
The downward motion is taken as positive and so, when the ball rises up, its
velocity is negative.
As the ball moves upwards,
the force of gravity (which is downwards) acts against its upward motion,
decreasing its velocity until it becomes zero. This is the maximum height
reached (at time = 0.50 s).
The area under a
velocity-time graph gives the displacement. So, to obtain the maximum height
reached we calculate the area under the graph between t = 0.30 s and t = 0.50
s.
Maximum height = area
under graph
Maximum height = ½ × (0.50
– 0.30) × 2.00 = 0.20 m
why was the height of triangle not 4?
ReplyDeletefrom 0.30s to 0.5s, the ball moves from the ground to the max height. So, we need to find this area under the graph.
DeleteOR from 0.5s to 0.7s, the ball falls from max height to ground. Alternatively, this area would also give the distance.
BUT not both together