Direct collision with a barrier at an angle

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Homework Help Overview

The problem involves a ball colliding with a barrier at an angle, focusing on the effects of the collision, including speed after impact, impulse, kinetic energy loss, and the angle of the new path relative to the horizontal. The subject area includes dynamics and vector resolution in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss vector components of the ball's velocity before and after impact, the calculation of impulse, and the loss of kinetic energy. There is a specific focus on the angle calculation in part (iv), with questions about the method of adding angles versus vectors.

Discussion Status

Participants are engaged in clarifying the correct approach to finding the angle after the collision. Some guidance has been offered regarding the addition of angles rather than vectors, and there is an acknowledgment of the potential for positive or negative angle interpretations.

Contextual Notes

There is a mention of the lack of specification regarding the positive or negative x-axis in the problem setup, which may affect the interpretation of the angle calculated.

Woolyabyss
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Homework Statement


A ball of mass m moves horizontally with speed 20 m/s towards a smooth barrier xy which makes an angle tan-inverse(4/3) with the horizontal.The coefficient of restitution is (3/4)
find

(i) The speed of the ball after the impact
(ii) The magnitude of the impulse due to the impact
(iii) the loss in kinetic energy
(iv) the tan of the angle between the ball's new path and the horizontal


Homework Equations





The Attempt at a Solution


let barrier xy = x axis
i = horizontal component vector
j = vertical component vector
(i)
resolving 20 m/s into component vectors 20(3/5) = 12i 20(4/5) = -16j
12i - 16j

i component remains the same

16(3/4) 12j

12i +12 j (changes direction) using Pythagoras theorem to find magnitude = 12√2 m/s

(ii)
since only the j vector changes

12m -(-16m ) = 28m Ns

(iii) (1/2)m(20^2) - (1/2)m(12√2)^2 = 56m j

(iv)
(12i - 16j) + (12i + 12j) = 24i - 4j j/i = tan∅ = 4/24 = 1/6

I got the right answers for part (i), (ii) and (iii)
(iv) is the part I am stuck on, is it not as simple as adding the two vectors and finding the angle? the answer at the back of my book is 7
 
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Woolyabyss said:
I got the right answers for part (i), (ii) and (iii)
(iv) is the part I am stuck on, is it not as simple as adding the two vectors and finding the angle? the answer at the back of my book is 7

No it is not that simple. You have got to add angles not Vectors.
 
darkxponent said:
No it is not that simple. You have got to add angles not Vectors.

Thanks I forgot that the ratio and the angle don't increase proportionally. Is this correct?
tan-inverse(4/3) +tan-inverse(12/12) = 98.1301

tan98.1301= -7
does it matter that its - 7?
can I just take it as positive?
 
Woolyabyss said:
Thanks I forgot that the ratio and the angle don't increase proportionally. Is this correct?
tan-inverse(4/3) +tan-inverse(12/12) = 98.1301

tan98.1301= -7
does it matter that its - 7?
can I just take it as positive?

They didn't specify whether the positive x-axis or the negative x axis. So answer you got is correct!
 
darkxponent said:
They didn't specify whether the positive x-axis or the negative x axis. So answer you got is correct!

Alright thanks.
 
Woolyabyss said:
Alright thanks.

You are welcome!
 

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