Two particles with masses m and 3m are moving toward each other along the x axis with the same initial speeds Vi. Particle m is traveling to the left, and the particle 3m is traveling to the right. They undergo an elastic glancing collision such that particles m is moving in the negative y direction after the collision at a right angle from its initial direction. (a) Find the final speeds of the two particles in terms of Vi. (b) What is the angle theta at which the particle 3m is scattered?

To solve this problem, we can use the principles of conservation of momentum and conservation of kinetic energy. Let's break down the problem into two parts:

(a) Finding the final speeds of the two particles in terms of Vi:

According to the principle of conservation of momentum, the initial momentum of the system is equal to the final momentum of the system. Since there are no external forces acting on the system, momentum is conserved.

Initially, particle m has momentum -mVi (negative because it is moving in the opposite direction) and particle 3m has momentum 3mVi.

After the collision, particle m is moving in the negative y direction at an angle of 90 degrees from its initial direction, which means it has no momentum in the x direction. Therefore, the x-component of its momentum is zero.

Let's denote the final speed of particle m as vf_m and the final speed of particle 3m as vf_3m.

Using conservation of momentum:
-mVi + 3mVi = 0 + m(vf_m)cos(90) + 3m(vf_3m)cos(theta) (Equation 1)

Now, let's consider conservation of kinetic energy. Since it is an elastic collision, kinetic energy is conserved.

Using conservation of kinetic energy:
(1/2)mVi^2 + (1/2)(3m)(Vi)^2 = (1/2)m(vf_m)^2 + (1/2)(3m)(vf_3m)^2 (Equation 2)

Simplifying equation 1:
mVi = 3m(vf_3m)cos(theta)

Simplifying equation 2:
mVi^2 + 3mVi^2 = m(vf_m)^2 + 3m(vf_3m)^2
4mVi^2 = m(vf_m)^2 + 3m(vf_3m)^2

Dividing both sides of equation 2 by m:
4Vi^2 = (vf_m)^2 + 3(vf_3m)^2

Now, we have two equations with two unknowns (vf_m and vf_3m), and we can solve them simultaneously.

(b) Finding the angle theta at which particle 3m is scattered:

The angle at which particle 3m is scattered can be found using the conservation of momentum equation 1.

-mVi + 3mVi = 0 + m(vf_m)cos(90) + 3m(vf_3m)cos(theta)

Simplifying:
-mVi + 3mVi = 3m(vf_3m)cos(theta)

Dividing both sides by 3m:
-vi + Vi = vf_3mcos(theta)

Simplifying further:
Vi - vf_3mcos(theta) = vi

Rearranging the equation:
vf_3mcos(theta) = Vi - vi

Dividing both sides by vf_3m:
cos(theta) = (Vi - vi) / vf_3m

Take the inverse cosine (arccos) of both sides to find theta:
theta = arccos((Vi - vi) / vf_3m)

In summary, to find the final speeds of the particles (vf_m and vf_3m) in terms of Vi and the angle theta at which particle 3m is scattered, we need to solve the system of equations derived from conservation of momentum and conservation of kinetic energy.