This is a suggested solution to the GCE O Level Physics examination in year 2016, Paper 2, Section A, question 1.
In this question, it was described that a student stood near a cliff and threw a ball upwards. The ball rises vertically for a short distance in the air and then falls. It was stated in the question that the air resistance is svery small as the ball moves upwards.
In this question, students are required to fill in the table on the magnitude and direction of the acceleration as the ball moves upwards and when it reaches the highest point of the motion.
The suggested answer to this part (a) of the question is as follows:
Magnitude of acceleration | Direction of acceleration | |
As the ball moves upward | 10 m/s2 | downwards |
At the highest point of the motion | 10 m/s2 | downwards |
(b) In the second part of the question, a displace-time graph for the first 1.0s of the motion was provided. It was stated in the question that air resistance is very small in the first 1.0s of the motion and students are required to sketch the velocity time graph for the first 1.0s of the motion.
To plot the velocity-time graph from the displacement-time graph, the standard way is to find the gradient at each specific point of the displacement-time graph. However, for this case, the questionÂ A on the given graph paper as the initial velocity. The question also state that air resistance is very small in the first 1.0s of the motion. Hence, the key concepts that the student must show in the answer are:
- Since air resistance is very small, considered as negligible for the first 1.0s, the acceleration due to gravity is therefore a constant. In a velocity-time graph, the gradient gives the acceleration. Hence if the acceleration is constant, then the graph must be a straight line.
- If the velocity-time graph is a straight line, then 2 points are needed for the graph to be sketched out. The first point, the initial velocity has been given as a point marked A. The second point is when the displacement reached the maximum, i.e. when the ball reached the maximum height. At maximum height, velocity is zero. With this 2 points the graph can be sketched as follows:
(c ) (i) In this part of the question, the student is required to state how the displacement-time graph (given in Fig 3.2 above) shows that the terminal velocity is not reached in the first 1.0s. The student should specify that Terminal Velocity is when air resistance equals the weight of the ball, causing the ball to fall at a constant velocity. Constant velocity on a displacement-time graph means the graph has to be a straight line. Since the displacement-time graph in the first 1.0s is a curve with no section that is a straight line, there was no constant velocity and hence we can conclude that terminal velocity were not reached in the first 1.0s.
(d) (ii) It was stated in the question that as the ball continues to fall, the effect of air resistance becomes significant and the ball eventually falls at terminal velocity. The student is required to describe the velocity and acceleration of the ball as it falls at terminal velocity. The suggested answer to this part of the question is —At terminal velocity, the ball’s velocity is constant and its acceleration is zero.
The graph background by Unknown Author is licensed under CC BY-SA-NC