Impulse/Vector question
https://ca99c64778b62ba7e7b339967029e090c5733c3a.googledrive.com/host/0B1ZiqBksUHNYOXdJSDFGUUZPRUE/June%202014%20(IAL)%20QP%20-%20M2%20Edexcel.pdf
Q5 part i. I understand the i and j direction equations in the ms, but cant understand how the (12+2K)^2 + (2K)^2 = 225 is set up. Solved this by obtaining the angle first, but I would like to know this less painful way.
Q5 part i. I understand the i and j direction equations in the ms, but cant understand how the (12+2K)^2 + (2K)^2 = 225 is set up. Solved this by obtaining the angle first, but I would like to know this less painful way.
Original post by shehab77
https://ca99c64778b62ba7e7b339967029e090c5733c3a.googledrive.com/host/0B1ZiqBksUHNYOXdJSDFGUUZPRUE/June%202014%20(IAL)%20QP%20-%20M2%20Edexcel.pdf
Q5 part i. I understand the i and j direction equations in the ms, but cant understand how the (12+2K)^2 + (2K)^2 = 225 is set up. Solved this by obtaining the angle first, but I would like to know this less painful way.
Q5 part i. I understand the i and j direction equations in the ms, but cant understand how the (12+2K)^2 + (2K)^2 = 225 is set up. Solved this by obtaining the angle first, but I would like to know this less painful way.
Find the new velocity in the vertical direction, then the new velocity in the horizontal direction.
The result velocity is 15. So vertical^2 + horizontal^2 = resultant^2.
Original post by Zacken
Find the new velocity in the vertical direction, then the new velocity in the horizontal direction.
The result velocity is 15. So vertical^2 + horizontal^2 = resultant^2.
The result velocity is 15. So vertical^2 + horizontal^2 = resultant^2.
Thanks! got it.
Original post by shehab77
Thanks! got it.
Nice!
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