Elastic Glancing Collision-HELP

[psychfan29]

Elastic Glancing Collision-HELP!

A white ball, mass of 1 kg has a speed of 1.68 m/s and a yellow ball, mass of 2kg, is at rest prior to an elastic glancing collision. After the collision the white ball has a speed of 1.24 m/s. To the nearest tenth of a degree, measured counterclockwise from east, what angle does it scatter at if the yellow ball is scattered at 280degrees?

I don't know what equations to use for this problem!

 

[psychfan29]

I used this equation:

1/2mv_1i^2 = 1/2mv_1f^2 + 1/2mv_2f^2 (the 1i, 1f, and 2f are subscripts indicating which velocities)

When I plugged in the numbers from the problem, I got that the final velocity for the yellow ball is 0.801m/s.

I then used the following equation:

Py=1kg(1.24m/s)sintheta+2kg(vyellow)sin280

But sin becomes greater than 1, so it's a domain error.

Am I doing something completely wrong?
Or is it possible that for this problem to work, the mass of the white ball must be greater than that of the yellow ball?

 

[baba89]

Hello! Thank you for reaching out for help with this problem. Let's break it down and see what equations we can use to solve it.

Firstly, we know that this is an elastic collision, which means that both momentum and kinetic energy are conserved. We can use the equations for conservation of momentum and conservation of kinetic energy to solve for the angle at which the white ball scatters.

Conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. In this case, we have two objects colliding, so we can write this as:

m1v1i + m2v2i = m1v1f + m2v2f

Where m1 and m2 are the masses of the white and yellow balls respectively, v1i and v2i are the initial velocities of the white and yellow balls, and v1f and v2f are the final velocities of the white and yellow balls.

We know the values for m1, v1i, and v1f, so we can plug those in and solve for v2f:

1kg(1.68 m/s) + 2kg(0 m/s) = 1kg(1.24 m/s) + 2kg(v2f)
v2f = 0.56 m/s

Next, we can use the conservation of kinetic energy equation, which states that the total kinetic energy before the collision is equal to the total kinetic energy after the collision. In this case, we can write it as:

(1/2)m1v1i^2 + (1/2)m2v2i^2 = (1/2)m1v1f^2 + (1/2)m2v2f^2

Again, we know the values for m1, v1i, and v1f, and we just solved for v2f, so we can plug those in and solve for v2i:

(1/2)(1kg)(1.68 m/s)^2 + (1/2)(2kg)(0 m/s)^2 = (1/2)(1kg)(1.24 m/s)^2 + (1/2)(2kg)(0.56 m/s)^2
v2i = 1.36 m/s

Now that we have the initial velocity of the yellow ball, we can use trigonometry