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# M3 Impulse Momentum question watch

1. I don't get the answers to these simple questions for some reason..

2. A cricket ball of mass 160 grams is thrown on to the ground with a velocity of 25 m/s at 40deg to the horisontal. Because of friction the total impulse fromm the \ground is at 10 deg to the vertical. The ball rebounds at 25 deg to the horizontal. Calculate the speed of the ball on the rebound , and the magnitude of the toal impulse.

WHat am i doing wrong?
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2. (Original post by gunmetalpanda)
I don't get the answers to these simple questions for some reason..

2. A cricket ball of mass 160 grams is thrown on to the ground with a velocity of 25 m/s at 40deg to the horisontal. Because of friction the total impulse fromm the \ground is at 10 deg to the vertical. The ball rebounds at 25 deg to the horizontal. Calculate the speed of the ball on the rebound , and the magnitude of the toal impulse.

WHat am i doing wrong?
I think you would change sine with cosine and reverse at the RHS
3. (Original post by ztibor)
I think you would change sine with cosine and reverse at the RHS
But why?
4. (Original post by gunmetalpanda)
But why?
Not sure what all of ztibor's comment means, but the angle of the impulse is relative to the vertical whereas the other angles are relative to the horizontal.

In addition the impulse will be going left of vertical in your diagram (friction resists motion), though it shouldn't effect your calculations.

Edit: Correction; it will effect your calculations.
5. It looks like the easiest plan will be to resolve parallel and perpendicular to the direction of the *impulse* (rather than the ground).
6. (Original post by ghostwalker)
Not sure what all of ztibor's comment means, but the angle of the impulse is relative to the vertical whereas the other angles are relative to the horizontal.

In addition the impulse will be going left of vertical in your diagram (friction resists motion), though it shouldn't effect your calculations.

Edit: Correction; it will effect your calculations.
Sorry I don't get it, I changed the diagram but I don't know what you mean about the first point. If there is an angle at which the impulse takes place there WILL be a vertical and horizontal component of it right? Same goes for the other angles? By resolving that way I get the correct value of v but I stays wrong :/
7. (Original post by gunmetalpanda)
Sorry I don't get it, I changed the diagram but I don't know what you mean about the first point. If there is an angle at which the impulse takes place there WILL be a vertical and horizontal component of it right? Same goes for the other angles? By resolving that way I get the correct value of v but I stays wrong :/
Bear in mind DFranklin's point, which does simplify calculations.

Setting that aside for now and going with what you've put in your original post.

Consider conservation of momentum in the horizontal direction.

25 cos40 - I sin 10 = v cos 25

And all those terms should be the first component in their corresponding vectors (or they should all be the second, doesn't matter which as long as they are all in the same position).
8. (Original post by gunmetalpanda)
But why?
Because at the left hand side in the Impulse vector you wrote the
vertical component as first, and bellow the horizontal as second.
But at the right hand side you wrote the horizontal component as first
in both velocity vectors. When the x axis of coordinate system is horizontal then these velocity vectors are correct just then change the components of Impulse vector each other. When the x axis of coordinate system is vertical thent the Impulse vector
is correct and change the velocity vectors components as I wrote.

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