Direct collision with a barrier at an angle

AI Thread Summary
A ball with a mass m collides with a barrier at an angle, initially moving at 20 m/s. After the impact, the speed of the ball is calculated to be 12√2 m/s, with an impulse of 28m Ns and a kinetic energy loss of 56m j. The angle between the ball's new path and the horizontal requires adding angles rather than vectors, leading to a calculated angle of approximately 98.13 degrees. The discussion clarifies that the sign of the angle does not affect the correctness of the answer. Overall, the calculations for parts (i), (ii), and (iii) are confirmed correct, while part (iv) emphasizes the importance of angle addition.
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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