Conservation of Momentum (Elastic Collision)

[tanzl]

Homework Statement

A particle with mass m and speed v collides with another particle with
mass m, which is initially stationary. The collision is elastic. Show that
the particles travel in perpendicular directions after the collision.

Homework Equations

1)initial momentum = final momentum
2)initial kinetic energy = final kinetic energy

The Attempt at a Solution

Assume that B is the moving particle and A is the stationary particle.

By conservation of momentum,
For the horizontal component of the motion
mv_i=mv_fcos[tex]\theta[/tex]+mu_fcos[tex]\phi[/tex]

For the vertical component of the motion
0=mv_fsin[tex]\theta[/tex]+mu_fsin[tex]\phi[/tex]

where
v_i = initial velocity of B
v_f = final velocity of B
u_f = final velocity of B
[tex]\theta[/tex] = angle between vector A and horizontal axis
[tex]\phi[/tex] = angle between vector B and horizontal axis

By conservation of energy,
[tex]\frac{1}{2}[/tex]mv_i²=[tex]\frac{1}{2}[/tex]m(v_f²+u_f²)
So, v_i²=v_f²+u_f²

These are all the equations (3 equations 5 variables) I can get,
But it seems that I have too many variable so I can't solve for [tex]\theta[/tex] and [tex]\phi[/tex]. Is there any more assumption I need to make in order to solve the question?

 

[LowlyPion]

Homework Statement

A particle with mass m and speed v collides with another particle with
mass m, which is initially stationary. The collision is elastic. Show that
the particles travel in perpendicular directions after the collision.

Homework Equations

1)initial momentum = final momentum
2)initial kinetic energy = final kinetic energy

The Attempt at a Solution

Assume that B is the moving particle and A is the stationary particle.

By conservation of momentum,
For the horizontal component of the motion
mv_i=mv_fcos[tex]\theta[/tex]+mu_fcos[tex]\phi[/tex]

For the vertical component of the motion
0=mv_fsin[tex]\theta[/tex]+mu_fsin[tex]\phi[/tex]

where
v_i = initial velocity of B
v_f = final velocity of B
u_f = final velocity of B
[tex]\theta[/tex] = angle between vector A and horizontal axis
[tex]\phi[/tex] = angle between vector B and horizontal axis

By conservation of energy,
[tex]\frac{1}{2}[/tex]mv_i²=[tex]\frac{1}{2}[/tex]m(v_f²+u_f²)
So, v_i²=v_f²+u_f²

These are all the equations (3 equations 5 variables) I can get,
But it seems that I have too many variable so I can't solve for [tex]\theta[/tex] and [tex]\phi[/tex]. Is there any more assumption I need to make in order to solve the question?

Looks like a good start. You look armed and dangerous.

 

[Clairefucious]

In these sorts of problems I find it useful to draw of diagram of the vectors. You know that the sum of all the x- and y-velocity vectors before and after the collison are equal which gives you the following options:

y-velocity:

uf*sin(phi) = - vf*sin(theta) Equation (1)
[equal and opposite final velocities of the two particles in the y-direction]

a) phi = 0, theta = 180 (in degrees)
b) phi = 180, theta = 0

c) 0 <= phi <= 90, -90 <= theta <= 0
d) 0 <= theta <= 90, -90 <= phi <= 0

From conservation of kinetic energy, you can eliminate options a) and b) for these particles of mass 'm'. You can then choose either c) or d) without loss of generality.

You then need to sub in Equation (1) to the equations you have for conservation of x-momentum and conservation of kinetic energy. It will require some funky algebra to solve the equation, but the trick is to place certain boundaries on what the final angles can be.

Hope this helps

 

[tanzl]

Thank you for your help.