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2D Collision : Glancing Collision

  • Thread starter n3w ton
  • Start date
  • #1
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Homework Statement


Two balls of equal mass m undergo a collision, the velocities of the balls make angles of 25.5° and -45.9° relative to the original direction of motion of the moving ball.
(a) Draw and label a diagram to show the balls before and after the collision. Label the angles (something like alpha and beta)
(b)Calculate the final speeds of the balls if the initial ball had a speed of 3.63 m/s.

[Answer:2.75 m/s and 1.65 m/s]


Homework Equations


Momentum Is Conserved in both X & Y Direction

X-Direction
m1vi1x + m2vi2x = m1vf1x + m2vf2x

Y-Direction
m1vi1y + m2vi2y = m1vf1y + m2vf2y

The Attempt at a Solution



(B)
X-DIRECTION
Mass 1= Mass 2= M

m1vi1x + m2vi2x = m1vf1x + m2vf2x
[STRIKE](m)[/STRIKE]vi1x + [STRIKE](m)[/STRIKE]vi2x = [STRIKE](m)[/STRIKE]vf1x + [STRIKE](m)[/STRIKE]vf2x
vi1x + vi2x = vf1x + vf2x
(3.63 m/s [forward]) + 0 m/s = vf1cos∅ + vf2cos∅
(3.63 m/s [forward]) = vf1cos∅ + vf2cos∅
3.63 m/s = vf1cos25.5° - vf2cos45.9° EQUATION 2

Y-DIRECTION
Mass 1= Mass 2= M

m1vi1y + m2vi2y = m1vf1y + m2vf2y
[STRIKE](m)[/STRIKE]vi1y + [STRIKE](m)[/STRIKE]vi2y = [STRIKE](m)[/STRIKE]vf1y + [STRIKE](m)[/STRIKE]vf2y
vi1y + vi2y = vf1y + vf2y
0 m/s + 0 m/s = vf1cos∅ + vf2cos∅
0 = vf1sin∅ + vf2sin∅
0 = vf1sin25.5° - vf2sin45.9°
vf2sin45.9° = vf1sin25.5°
vf2sin45.9° / sin25.5° = vf1 EQUATION 1

SUBSTITUTION 1 INTO 2

3.63 = [vf2sin45.9° / sin25.5°] cos25.5° - vf2cos45.9°



===================

Now I would have to Re-Arrange one of them and for either vf1 or vf2 and substitute, but when I do it I get a wrong answer (I think I may be entering the angles wrong)

Can Someone please do this question and check if they get the answers they should of Answer:2.75 m/s and 1.65 m/s
 
Last edited:

Answers and Replies

  • #2
gneill
Mentor
20,801
2,778
0 = vf1sin25.5° + vf2sin45.9°
0 = vf1sin25.5° - vf2sin45.9° <---Would this be negative?
The angle in the problem statement is given as -45.9°, so you can use the positive angle if you 'promote' the sign out of the sin() function and apply it to the term as you've done in the second line above. (The previous line is technically incorrect because you've forgotten the negative sign on the angle)

You'll have to show more of the algebra you're doing from this point if we are to see what's going wrong with your attempt to solve for the two unknowns.
 
  • #3
19
0
Here:

SUBSTITUTION 1 INTO 2

3.63 = [vf2sin45.9° / sin25.5°] cos25.5° - vf2cos45.9°
3.63 = 1.5055[vf2 - 0.6959vf2
3.63/(1.5055-0.6959) = vf2
4.48 m/s = Vf2
 
  • #4
gneill
Mentor
20,801
2,778
Here:

SUBSTITUTION 1 INTO 2

3.63 = [vf2sin45.9° / sin25.5°] cos25.5° - vf2cos45.9°
3.63 = 1.5055[vf2 - 0.6959vf2
3.63/(1.5055-0.6959) = vf2
4.48 m/s = Vf2
Why is the term in red negative? You can extract the sign of the angle from the sin() function for the given angle of -45.9°, but you can't do that for the cos() function; the cosine of -45.9° is a positive value, and is in fact the same as the cosine of 45.9°.
 

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