# Maximum Deflection Angle In Elastic Collision

• vibha_ganji
In summary, the author is trying to solve for cos theta but is having trouble. He suggests using LaTeX to make the equations easier to read.

#### vibha_ganji

Homework Statement
Show that, in the case of an elastic collision of a particle of
mass m1 with a particle of mass m2 , initially at rest, (a) the
maximum angle m through which m1 can be deflected by the
collision is given by cos^2(theta) = 1 - (m2/m1)^2 so that
0 is less than or equal to theta which is less than or equal to pi/2 and when when m1 < m2.
Relevant Equations
v1f = ((m1-m2)/(m1+m2))*(v1i)
v of center of mass reference frame = (m1v1i + m2v2i)/(m1+m2)
I started by writing the equation v1i + v1f = 2v and then drawing a triangle with v1i, v1f, and 2v as the three sides. Then I used the Law of Cosines to solve for cos theta but this did not lead to a solution. Could I have a hint on how to begin? Thank you!

vibha_ganji said:
I started by writing the equation v1i + v1f = 2v
Why? How are you defining those variables, and by what reasoning do you get that equation?

The variable v1i is the initial velocity of mass m1 in the laboratory frame and v1f is the final velocity of mass m1 in the laboratory frame.The variable v is the difference in velocity between the center of mass frame and the laboratory frame. Therefore, we can write that the initial velocity of mass m1 in the center of mass reference frame is v1i’ = v1i - v. Similarly, we can write that v1f’ = v1f - v. Since it is an elastic collision, v1i’ is the negation of v1f’. We can substitute-(v1i-v) for v1f’ and write it as -v1i+v = v1f - v. This can be simplified to v1i + v1f = 2v.

vibha_ganji said:
Since it is an elastic collision, v1i’ is the negation of v1f’.
No, that is only true for the components of velocity normal to the contact plane.
Notice that since m2 is initially stationary v1i and v are parallel. Your equation leads to v1f also being parallel to these.

Oh ok that makes more sense! Could I have a hint on how to continue?

vibha_ganji said:
Oh ok that makes more sense! Could I have a hint on how to continue?
The next step is to consider that this collision is two-dimensional and conserve momentum in two mutually perpendicular directions. The usual convention is to take the x-axis along the direction of motion of the projectile mass m1.

An auxiliary step would be for you to learn how to use LaTeX and use it to write your equations. It will make your equations easier to read and our advice easier to give.