What is the value of e (coefficient of restitution) for the following?

[Kaushik]

Homework Statement

Find e if the initial velocity, $u$ , of the ball is $40 \frac{m}{s}$ at an angle 30 degree with the horizontal. It collides with the wall which is $80m$ away. Find e if the ball returns through the same path.

Relevant Equations

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This is the setup.

My attempt:

The ball must have the same velocity before collision as it has just after the collision as it has to return through the same path. Hence $V_{approach} = V_{separation} \implies e = 1$. Is it correct? If not, why and how should I approach this question?

 

[BvU]

Correct
[edit] e=1 for elastic is correct. Didn't look any further

 

[PeroK]

Homework Statement: Find e if the initial velocity, $u$ , of the ball is $40 \frac{m}{s}$ at an angle 30 degree with the horizontal. It collides with the wall which is $80m$ away. Find e if the ball returns through the same path.
Homework Equations: .

View attachment 252550
This is the setup.

My attempt:

The ball must have the same velocity before collision as it has just after the collision as it has to return through the same path. Hence $V_{approach} = V_{separation} \implies e = 1$. Is it correct? If not, why and how should I approach this question?

This looks very odd to me. In order to return by the same path the ball would have to strike the wall horizontally. Does it reach the wall just at its highest point?

If not, then the wall would need to be tilted at just the correct angle to reflect the ball back the way it came.

Or, the ball would get a strange bounce - again at just the perfect angle to send it back the way it came.

The question doesn't look right to me.

 

[Kaushik]

Does it reach the wall just at its highest point?

Here, the height of the wall is not mentioned.

Also, why should the wall have a specific angle for it to return through the same path? For different $e$ it follows different path. They are asking us to find that value of $e$ such that it returns through the same path. Isn't? Am I misinterpreting something here?

 

[haruspex]

Homework Statement: ... the initial velocity, $u$ , of the ball is $40 \frac{m}{s}$ at an angle 30 degree with the horizontal. It collides with the wall which is $80m$ away. ... the ball returns through the same path.

How is this possible? The wall is beyond the peak of the trajectory.
Either the wall is at an angle or the ball had top spin.

 

[Kaushik]

How is this possible? The wall is beyond the peak of the trajectory.
Either the wall is at an angle or the ball had top spin.

What if the distance from the initial point to the wall is $30m$?

 

[Kaushik]

This looks very odd to me. In order to return by the same path the ball would have to strike the wall horizontally. Does it reach the wall just at its highest point?

If not, then the wall would need to be tilted at just the correct angle to reflect the ball back the way it came.

Or, the ball would get a strange bounce - again at just the perfect angle to send it back the way it came.

The question doesn't look right to me.

(aha moment ) Oh I just realized what you actually meant. Makes sense!
So either the value $80m$ or that the wall is vertical is wrong. What will we get if we consider wall to be at a distance of $30m$ away?

 

[PeroK]

(aha moment ) Oh I just realized what you actually meant. Makes sense!
So either the value $80m$ or that the wall is vertical is wrong. What will we get if we consider wall to be at a distance of $30m$ away?

Angle of incidence = angle of reflection for an elastic collision.

In any case, for the ball to rebound back the way it came the angle of incidence must be a right angle. Unless you have spin or the wall is uneven. With a brick wall, for example, the ball could strike the indentations between the bricks and come off at an unusual angle. But, I don't imagine we are considering that sort of thing here.

 

[Kaushik]

Angle of incidence = angle of reflection for an elastic collision.

In any case, for the ball to rebound back the way it came the angle of incidence must be a right angle. Unless you have spin or the wall is uneven. With a brick wall, for example, the ball could strike the indentations between the bricks and come off at an unusual angle. But, I don't imagine we are considering that sort of thing here.

The question is not that complicated, i.e. we are not considering the wall to be uneven.

• If the wall is considered to be vertical then the ball should collide when it is at $H_{max}$. So the distance of the wall from the initial point must be different (closer to the initial point). Isn't?
• If not, then the wall should be inclined such that the ball collides orthogonal to the wall. Isn't?

 

[PeroK]

What if the distance from the initial point to the wall is $30m$?

Angle of incidence = angle of reflection for an elastic collision.

The question is not that complicated, i.e. we are not considering the wall to be uneven.

• If the wall is considered to be vertical then the ball should collide when it is at $H_{max}$. So the distance of the wall from the initial point must be different (closer to the initial point). Isn't?
• If not, then the wall should be inclined such that the ball collides orthogonal to the wall. Isn't?

Yes. You could study those cases out of interest. But, I can't say what the questioner intended here.

 

[Kaushik]

Angle of incidence = angle of reflection for an elastic collision.Yes. You could study those cases out of interest. But, I can't say what the questioner intended here.

I think as of now, I'll discard this question due to its ambiguity .