Ball's motion after one bounce off the ground - SUVAT Question (ENGAA 2017)

[TomK]

Homework Statement

ENGAA 2017 - Question 54

Relevant Equations

SUVAT

Please scroll-down to the end (Question 54): https://www.undergraduate.study.cam.ac.uk/files/publications/engineering_s1_qp_2017.pdf

I have also been referring to unofficial worked solutions (http://www.engineeringadmissionsassessment.com/2017-solutions.html), but I didn't understand how it gives the answer.

The correct answer is 'A' (speed = 4, height = 0.8). This is the working I did:

u = 0, v = ?, t = 0.5, a = 10

v = u + at = 0 + (10 x 0.5) = 5: speed = 5

From here, I said the starting speed immediately after the first impact is 5 (and the speed at max. height = 0), but, depending on the equation/information I use, I seem to get two different height values:

u = 5, s = ?, a = -10, t = 0.4.

s = ut + (1/2)at² = 2 + (-5 x (0.4)²) = 2 - 0.8
s = 1.2m

Or: u = 5, s = ?, a = -10, v = 0

v² = u² + 2as: s = (v² - u²)/2a = -25/-20
s = 1.25m

Either way, it gives me the wrong answer of 'D' (speed = 5, height = 1.25).

Would someone be able to tell me what has gone wrong here? Am I not allowed to assume the speed is the same after hitting the ground?

 

[Merlin3189]

Homework Statement:: ENGAA 2017 - Question 54
For the first fall, starting from rest
u = 0, v = ?, t = 0.5, a = 10

v = u + at = 0 + (10 x 0.5) = 5: speed = 5

From here, I said the starting speed immediately after the first impact is 5 (and the speed at max. height = 0),
This is where I'd disagree with you. I don't think you can assume a perfectly elastic bounce.

but, depending on the equation/information I use, I seem to get two different height values:

u = 5, s = ?, a = -10, t = 0.4.

s = ut + (1/2)at² = 2 + (-5 x (0.4)²) = 2 - 0.8
s = 1.2m
Which is the height it reaches after 0.4 sec after a perfectly elastic bounce. (*1)

Or: u = 5, s = ?, a = -10, v = 0

v² = u² + 2as: s = (v² - u²)/2a = -25/-20
s = 1.25m
Which is the height it actually reaches after a perfectly elastic bounce. (*2)
[/QUOTE]

*1 You use the speed of 100% bounce, but the time from the graph (not 100% bounce.)
*2 You use the speed of 100% bounce and find the max height it reaches.

 

[Steve4Physics]

Homework Statement:: ENGAA 2017 - Question 54
Relevant Equations:: SUVAT

Please scroll-down to the end (Question 54): https://www.undergraduate.study.cam.ac.uk/files/publications/engineering_s1_qp_2017.pdf

I have also been referring to unofficial worked solutions (http://www.engineeringadmissionsassessment.com/2017-solutions.html), but I didn't understand how it gives the answer.

The correct answer is 'A' (speed = 4, height = 0.8). This is the working I did:

u = 0, v = ?, t = 0.5, a = 10

v = u + at = 0 + (10 x 0.5) = 5: speed = 5

From here, I said the starting speed immediately after the first impact is 5 (and the speed at max. height = 0), but, depending on the equation/information I use, I seem to get two different height values:

u = 5, s = ?, a = -10, t = 0.4.

s = ut + (1/2)at² = 2 + (-5 x (0.4)²) = 2 - 0.8
s = 1.2m

Or: u = 5, s = ?, a = -10, v = 0

v² = u² + 2as: s = (v² - u²)/2a = -25/-20
s = 1.25m

Either way, it gives me the wrong answer of 'D' (speed = 5, height = 1.25).

Would someone be able to tell me what has gone wrong here? Am I not allowed to assume the speed is the same after hitting the ground?

You can’t assume a bounce is perfectly elastic. Some energy is lost on each impact.

This means the max. height of the ball after the first bounce is less than the height the ball is dropped from.

And the speed just after an impact is less than the speed just before the impact.

You have to ignore what happens between t=0 and t=0.5s.

From t=0.5s to t=1.3s the ball goes up and comes back down. Total time taken is 1.3 – 0.5 = 0.8s with max. height in the middle, which is 0.8/2 = 0.4s after the 1st impact. The acceleration is -10m/s² in flight.

If you take the start of the bounce (after 0.5s) as t=0 and unknown initial velocity (u), you can solve the problem.

 

[TomK]

You can’t assume a bounce is perfectly elastic. Some energy is lost on each impact.

This means the max. height of the ball after the first bounce is less than the height the ball is dropped from.

And the speed just after an impact is less than the speed just before the impact.

You have to ignore what happens between t=0 and t=0.5s.

From t=0.5s to t=1.3s the ball goes up and comes back down. Total time taken is 1.3 – 0.5 = 0.8s with max. height in the middle, which is 0.8/2 = 0.4s after the 1st impact. The acceleration is -10m/s² in flight.

If you take the start of the bounce (after 0.5s) as t=0 and unknown initial velocity (u), you can solve the problem.

Thank you. I now get u=4 (using v = u + at) and s=0.8 (using s = ut + (1/2)at²).

 

[Steve4Physics]

Thank you. I now get u=4 (using v = u + at) and s=0.8 (using s = ut + (1/2)at²).

Well done. On a matter of 'style', don't forget units. u = 4m/s and s = 0.8m. Forgetting units can lose a mark in an exam'. Generally, it creates a poor impression and can lead to errors if someone else is using your result and assumes the wrong units.