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Wonders of Physics


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During impact of a free falling ball, the force on ground is greater than the weight of ball

In this post, it shows a free-falling ball from a height of 1.0 m. During the impact, the direction of the force on the ground is downwards and the force on the ground by the ball is greater then the weight.

As the ball is free-falling, the only force acting is its weight downwards. Hence a common misconception is to think that the force on the ground during impact is equal to the weight. This is wrong.

The normal force (force on the ball by the ground = stopping force on the ball by the ground) is greater than the weight.

The force on the ball by the ground is equal and opposite to the force on the ground by the ball. Hence the magnitude of the force on the ground is greater than the weight.

Similar concept can be applied if a man jumps off from a height. But in this case, the man’s leg will exert a stopping force over a short distance. That stopping force, once again, is greater than the weight of the man.


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Man Jumps Vertically Upwards, Pressure On Ground Is Greater During The Jump

This concept is similar to a 2016 O-Level Pure Physics Question P2 Q2, on why the pressure acting on the ground is greater during the jump, compared to when he is standing stationary on the ground.

During the jump, his leg will exert an upward force. This upward force (equivalent to normal force or force on the man by the ground) is greater than the weight of the man. Hence there is a net (resultant force) upwards, causing him to accelerate upwards.

That force on the man by the ground is equal and opposite to the force on the ground by the man. This is an action-reaction pair. Since the force exerted on the ground by the man is greater (greater than weight), the pressure exerted on the floor is greater.

(NOTE: Normal force and Weight is not an action-reaction pair)


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2017PurePhyP2Q5 Two points on the rope wave with displacement-time graphs given

In this this question, the displacement-time graphs are given, which are different from displacement-distance graphs.

In the displacement-time graphs of A and B, they show the displacement of that particular point at different timing. E,g, at t = 0s, the A is at the rest position (0 displacement) and at time 0.2 s it is at the maximum displacement. This means A is going up from t = 0 s to 0.2 s.

Solutions:
(a) Amplitude: 1.5 cm
(b)(i) Frequency is the number of complete waves produced in 1 second.
(ii) period T = 0.8s, f = 1/T = 1/0.8 = 1.25 Hz
(c) Closest possible positions of A and B, (refer to the video), is when the
time taken for the wave to move from A to B is T/4 = 0.8/4 = 0.2 s.
speed = distance/time = 38/0.2 = 190 cm/s approx. 200 cm/s
(ii) There are various possibilities in which B can be 38 cm to the right of A. Besides T/4, it can be 1.25T or 2.25 T etc. Hence the speed can be other values.

Refer to the video for the full explanation