00:02
The problem states a ball with mass 0 .590 kilogram.
00:09
Let's place it as m1, 0 .590 kilograms.
00:14
It is struck by a second ball having a mass of 0 .390 kilograms.
00:25
Initially moving with a velocity of 0 .275 meters per second toward the right along the x axis.
00:36
So we have m2 moving to the right and striking the m1 ball.
00:46
So m1 is said to be at rest, hence b1 is equal to 0.
00:54
After the collision, the 0 .390 kilogram ball has a velocity of 0 .185 meters per second at an angle of 37 .8 degrees above the x -axis.
01:09
So if this is the m2 ball, the angle here is 37 .8 degrees.
01:21
Now, the resulting direction of the m1 ball should be directed downward, but the angle is still unknown.
01:34
So you have theta 2 and theta 1, m1.
01:40
The final velocity of the mass 2 after collision, so let's place it as v2 prime, is equal to 0 .185 meters per second.
01:56
And then the final velocity of the v1 ball or the m1 ball is also unknown.
02:06
Now, both balls move on a frictionless horizontal surface, and if that is the case, what is the magnitude of? of the velocity of m1 after the collision.
02:22
In collision problems, momentum is always conserved, and that means the summation of momentum in the initial condition should be equal to the summation of momentum in the final condition.
02:36
And that can be further divided into its components, or the component of momentum along x, and the component of momentum along y.
02:52
So starting with the component of x, we'll have the initial condition of m2 moving to the right with the velocity v2 and then m1 at rest.
03:06
Then the final conditions would have m2 moving at a new velocity with an angle of theta 2, and then m1 also moving at a new velocity with an angle of theta 1.
03:28
Now, going to the y component, let's place it here.
03:34
There is no y component of momentum in the initial condition, since the only movement is along the x.
03:42
Hence, we'll only have the terms in the final momentum, which is m2v2 prime, sine theta, since we're getting the y component now, minus m1v1 prime, sine theta 1 since the mass 1 or the m1 ball should be moving downward after the collision so in order to solve for the unknown which is v1 you need to see that the first equation we have here has two unknowns the unknown being the angle of theta 1 and then it's its velocity of v1.
04:42
And then for the second equation or the momentum along the y, we also have two unknowns, v1 and then your theta 1.
04:52
So we have two unknowns and two equations.
04:56
You could easily solve this using systems of linear equation.
05:02
So isolating v1 from our second equation will have v1 prime is equal to m2 v2 prime sine theta 2 divided by m1 sine theta 1 so it just places term to the left side of the equation and then divide it by m1 and sine theta 1 so substituting this equation of v1 prime will have a new equation of vm 2 v2 v2 equals a m2v2 prime, cosine theta 2 plus m1 times m2 v2 prime, sine theta 2 divided by m1, sine theta 1, sine theta 1.
06:12
So we cancel m1 here, and we could replace sine theta 1 and cosine here.
06:21
As a tangent function as tangent theta is equal to sine theta over cosine theta.
06:32
So we have m2 v2 equals m2 v2 prime cosine theta 2 plus m2 v2 prime sine theta 2 divided by tangent theta 1.
06:50
So all terms have m2.
06:53
It could cancel that out...