2D Collision : Glancing Collision

[n3w ton]

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
m₁v_i1x + m₂v_i2x = m₁v_f1x + m₂v_f2x

Y-Direction
m₁v_i1y + m₂v_i2y = m₁v_f1y + m₂v_f2y

The Attempt at a Solution

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

m₁v_i1x + m₂v_i2x = m₁v_f1x + m₂v_f2x
[STRIKE](m)[/STRIKE]v_i1x + [STRIKE](m)[/STRIKE]v_i2x = [STRIKE](m)[/STRIKE]v_f1x + [STRIKE](m)[/STRIKE]v_f2x
v_i1x + v_i2x = v_f1x + v_f2x
(3.63 m/s [forward]) + 0 m/s = v_f1cos∅ + v_f2cos∅
(3.63 m/s [forward]) = v_f1cos∅ + v_f2cos∅
3.63 m/s = v_f1cos25.5° - v_f2cos45.9° EQUATION 2

Y-DIRECTION
Mass 1= Mass 2= M

m₁v_i1y + m₂v_i2y = m₁v_f1y + m₂v_f2y
[STRIKE](m)[/STRIKE]v_i1y + [STRIKE](m)[/STRIKE]v_i2y = [STRIKE](m)[/STRIKE]v_f1y + [STRIKE](m)[/STRIKE]v_f2y
v_i1y + v_i2y = v_f1y + v_f2y
0 m/s + 0 m/s = v_f1cos∅ + v_f2cos∅
0 = v_f1sin∅ + v_f2sin∅
0 = v_f1sin25.5° - v_f2sin45.9°
v_f2sin45.9° = v_f1sin25.5°
v_f2sin45.9° / sin25.5° = v_f1 EQUATION 1

SUBSTITUTION 1 INTO 2

3.63 = [v_f2sin45.9° / sin25.5°] cos25.5° - v_f2cos45.9°



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

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

 

[gneill]

0 = v_f1sin25.5° + v_f2sin45.9°
0 = v_f1sin25.5° - v_f2sin45.9° <---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.

 

[n3w ton]

Here:

SUBSTITUTION 1 INTO 2

3.63 = [v_f2sin45.9° / sin25.5°] cos25.5° - v_f2cos45.9°
3.63 = 1.5055[v_f2 - 0.6959v_f2
3.63/(1.5055-0.6959) = v_f2
4.48 m/s = Vf2

 

[gneill]

Here:

SUBSTITUTION 1 INTO 2

3.63 = [v_f2sin45.9° / sin25.5°] cos25.5° - v_f2cos45.9°
3.63 = 1.5055[v_f2 - 0.6959v_f2
3.63/(1.5055-0.6959) = v_f2
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°.

 

[asteriatic]

I would approach this problem by first drawing a diagram to visualize the situation. I would label the angles alpha (α) and beta (β) as shown in the problem statement. From the diagram, it is clear that the initial velocity of the first ball can be broken down into its x and y components as 3.63 m/s cos α and 3.63 m/s sin α, respectively. Similarly, the initial velocity of the second ball can be broken down into its x and y components as 0 m/s cos β (since it is initially at rest) and 0 m/s sin β, respectively.

Using the equations for conservation of momentum in both the x and y directions, I would set up the following equations:

X-DIRECTION:
m(3.63 m/s cos α) + m(0 m/s cos β) = m(vf1 cos α) + m(vf2 cos β)

Y-DIRECTION:
m(3.63 m/s sin α) + m(0 m/s sin β) = m(vf1 sin α) + m(vf2 sin β)

Since we know the masses are equal (m1 = m2 = m), we can simplify the equations to:

X-DIRECTION:
3.63 m/s cos α = vf1 cos α + vf2 cos β

Y-DIRECTION:
3.63 m/s sin α = vf1 sin α + vf2 sin β

Next, we can use the trigonometric identities sin2α + cos2α = 1 and sin2β + cos2β = 1 to rewrite the equations as:

X-DIRECTION:
3.63 m/s = vf1 + vf2 cos(β-α)

Y-DIRECTION:
0 = vf1 sin(β-α) + vf2 sin(β+α)

We can now use substitution to solve for the final velocities. From equation 1, we can solve for vf1 in terms of vf2:

vf1 = 3.63 m/s - vf2 cos(β-α)

Substituting this into equation 2, we get:

0 = [3.63 m/s - vf2 cos(β-α)] sin(β-α) + vf2 sin(β+α)

Expanding and simplifying, we get:

0 = 3.63 m/s sin(β-α) - vf2 sin2(