Solving Collisions & ZMFs Homework: Mass 4 Particle Deflection

[villiami]

Homework Statement

A particle of relative mass 4 travels at velocity v, and collides with a stationary particle with relative mass 1. By considering the zero momentum frame, show that the larger particle can be deflected by an angle of arcsin(1/4) at the most.

(Note this is a non-relativistic problem)

Homework Equations

The Attempt at a Solution

 

[rl.bhat]

The problem can be solved by using conservation energy and momentum.
Lat v1i is the initial velocity of mass 4 and v2i is the initial velocity of mass 1. You can wright three equations.
4v1i + 0 = 4v1f*cos(theta1) + v2f*cos(theta2) [ moments along the x direction]...(1)
4v1f*sin(theta1) = v2f*sin(theta2)[y-components of momentum]...(2)
0.5*4*v1i^2 = 0.5*4*v1f^2 + 0.5*v2f^2 [conservation of energy}...(3)
Now rewright equation(1) as 4v1i - 4v1f*cos(theta1) = v2f*cos(theta2) and square it. Now square equation (2) and add it to the above equation.After simplification you will get
5v1f^2 -8v1i*v1f*cos(theta1) + 3v1i^2 = 0. Here v1i is constant. For real root of this quadritic equation we must have [64v1i^2cos^2(theta) -4*5*3v1i^] > 0
[[64v1i^2{1 - sin^2(theta)} -60v1i^] > 0 Taking 4v1i^2 common we get
16 - 16sin^2(theta) - 15 > 0 or 1 - 16sin^2(theta) > 0 or sin(theta) < 1/4. That is the required result.

 

[ogb p]

To solve this problem, we can use the conservation of momentum and energy principles. In the zero momentum frame, the total momentum of the system before and after the collision is zero. This means that the velocity of the larger particle after the collision must be equal in magnitude but opposite in direction to the velocity of the smaller particle before the collision.

Using the conservation of energy, we can also determine the velocity of the larger particle after the collision. Since the total energy of the system is conserved, we can equate the kinetic energy of the particles before and after the collision:

1/2 * m1 * v^2 = 1/2 * m2 * v'^2

Where m1 is the mass of the smaller particle, v is its initial velocity, m2 is the mass of the larger particle, and v' is its velocity after the collision.

Solving for v', we get:

v' = v * √(m1/m2)

Substituting the given values of m1 = 1 and m2 = 4, we get:

v' = v * 1/2

This means that the velocity of the larger particle after the collision is half of the velocity of the smaller particle before the collision.

To determine the maximum angle of deflection, we can use trigonometry. In the zero momentum frame, the larger particle will be deflected by an angle θ, where:

sinθ = v'/v = 1/2

Solving for θ, we get:

θ = arcsin(1/2) = arcsin(1/4)

Therefore, the larger particle can be deflected by an angle of arcsin(1/4) at the most in the zero momentum frame. This also means that the maximum deflection angle in the laboratory frame will be less than arcsin(1/4) since the larger particle will have some momentum in the laboratory frame.