stoneroses38383
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The question goes a smooth ball of mass 2kg moving in the xy plane collides with a vertical wall containing the line y=x. Immidiately before the collision, it moves with velocity (4i +2j) ms^-1 and given that e = 1/3, find the velocity of the ball after collision.

Yes I know theres a thread for this question already but i honestly didnt understand the method. Heres what I did

i used the unit vector parallel to the plane (i+j)/sqrt(2)
i then dotted this to (4i +2j) to find the angle between the velocity vector and the plane -> I ended up with cos (theta) = 3sqrt(10)/10

Then to find the velocity paralell to the plane i used (4i+2j)costheta which didnt get me the correct velocity paralell to the plane which is (3i+3j)

any help?
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mqb2766
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(Original post by stoneroses38383)
The question goes a smooth ball of mass 2kg moving in the xy plane collides with a vertical wall containing the line y=x. Immidiately before the collision, it moves with velocity (4i +2j) ms^-1 and given that e = 1/3, find the velocity of the ball after collision.

Yes I know theres a thread for this question already but i honestly didnt understand the method. Heres what I did

i used the unit vector parallel to the plane (i+j)/sqrt(2)
i then dotted this to (4i +2j) to find the angle between the velocity vector and the plane -> I ended up with cos (theta) = 3sqrt(10)/10

Then to find the velocity paralell to the plane i used (4i+2j)costheta which didnt get me the correct velocity paralell to the plane which is (3i+3j)

any help?
You don't really need to find the angle as you solve for it then sub it back in. The dot product projects the velocity vector onto the wall (unit vector) to give its magnitude in the direction of the wall.
Magnitude of the velocity parallel to wall = (4i+2j).(i+j)/sqrt(2) = 6/sqrt(2)
To find the vector, multiply the magnitude by the wall's unit vector.
Last edited by mqb2766; 3 weeks ago
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stoneroses38383
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(Original post by mqb2766)
You don't really need to find the angle as you solve for it then sub it back in. The dot product projects the velocity vector onto the wall (unit vector) to give its magnitude in the direction of the wall.
Magnitude of the velocity parallel to wall = (4i+2j).(i+j)/sqrt(2) = 6/sqrt(2)
To find the vector, multiply the magnitude by the wall's unit vector.
i understand that all apart from the last line, why do we multiply magnitude by unit vector to find the vector paralell?
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mqb2766
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(Original post by stoneroses38383)
i understand that all apart from the last line, why do we multiply magnitude by unit vector to find the vector paralell?
A vector (basically polar form if you've come across that) is
magnitude * direction
where the direction is the unit vector.
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stoneroses38383
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(Original post by mqb2766)
A vector (basically polar form if you've come across that) is
magnitude * direction
where the direction is the unit vector.
LOL thank you so much
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stoneroses38383
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(Original post by mqb2766)
You don't really need to find the angle as you solve for it then sub it back in. The dot product projects the velocity vector onto the wall (unit vector) to give its magnitude in the direction of the wall.
Magnitude of the velocity parallel to wall = (4i+2j).(i+j)/sqrt(2) = 6/sqrt(2)
To find the vector, multiply the magnitude by the wall's unit vector.
just one more question, why cant i use multiply (4i+2j) by cos theta to get the horizontal velocity vector?
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mqb2766
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(Original post by stoneroses38383)
just one more question, why cant i use multiply (4i+2j) by cos theta to get the horizontal velocity vector?
Because its wrong :-)
4 is the horizontal velocity - dot with i
2 is the vertical velocity - dot with j
You do know this ...

You could multiply the scalar 6/sqrt(2) by cos(theta) though.
Last edited by mqb2766; 3 weeks ago
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stoneroses38383
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(Original post by mqb2766)
Because its wrong :-)
4 is the horizontal velocity - dot with i
2 is the vertical velocity - dot with j
You do know this ...

You could multiply the scalar 6/sqrt(2) by cos(theta) though.
ahhh i see, thanks so much i honestly cant thank you enough
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