Angle of particles after elastic collision

[ajl1989]

Homework Statement

A particle of mass M1 collides elastically with a particle of mass M2 at rest. Show that if M1>M2 then the angle of M1 after the collision cannot exceed the value arcsin(M2/M1).

Homework Equations

conservation of momentum: M1v₁=M1v₁`+M2v<sub>2</sub>`
conservation of kinetic energy: (1/2)M1v₁²=(1/2)M1v₁`<sup>2</sup>+(1/2)M2v<sub>2</sub>`²

The Attempt at a Solution

I've figured out that the velocities after collision are:
v₁`=v<sub>1</sub>*[(M1-M2)/(M1+M2)]
and
v<sub>2</sub>`=v₁*[(2M1)/(M1+M2)]
but I don't know how to find the maximum angle of M1 after collision. Please help

 

[Pythagorean]

Homework Statement

A particle of mass M1 collides elastically with a particle of mass M2 at rest. Show that if M1>M2 then the angle of M1 after the collision cannot exceed the value arcsin(M2/M1).

Homework Equations

conservation of momentum: M1v₁=M1v₁`+M2v<sub>2</sub>`
conservation of kinetic energy: (1/2)M1v₁²=(1/2)M1v₁`<sup>2</sup>+(1/2)M2v<sub>2</sub>`²

The Attempt at a Solution

I've figured out that the velocities after collision are:
v₁`=v<sub>1</sub>*[(M1-M2)/(M1+M2)]
and
v<sub>2</sub>`=v₁*[(2M1)/(M1+M2)]
but I don't know how to find the maximum angle of M1 after collision. Please help

what's the critical point between when M1 > M2 is true, and when it's no longer true?

what will M1 - M2 be at that point?

to get angle into your equation, think projection.

 

[nlightner]

!

I would first like to commend you for your efforts in solving this problem and for seeking assistance when needed. Let me provide some guidance on how to find the maximum angle of M1 after the collision.

First, we know that the angle of M1 after the collision can be represented by the inverse trigonometric function of the ratio of the perpendicular and hypotenuse sides of the triangle formed by the velocities v1 and v1`. This can be written as:

θ = sin^-1(v1`/v1)

Substituting the expressions for v1` and v1 from the conservation of momentum equation, we get:

θ = sin^-1[(v1*[(M1-M2)/(M1+M2)])/(v1)]

Simplifying this expression, we get:

θ = sin^-1[(M1-M2)/(M1+M2)]

Now, in order to find the maximum value of θ, we can use the properties of inverse trigonometric functions. In particular, the maximum value of θ for the inverse sine function is equal to the inverse sine of 1, which is equal to π/2 radians or 90 degrees.

Therefore, the maximum angle of M1 after the collision cannot exceed the value arcsin(M2/M1), which is equal to π/2 radians or 90 degrees. This makes intuitive sense as well, since if M1 is much larger than M2, the angle of M1 after the collision would be almost parallel to the initial direction of motion and cannot exceed 90 degrees.

I hope this helps you understand the concept better and solve the problem. Keep up the good work!