Two particles in an elastic collision given angle of first find other.

[oddjobmj]

Homework Statement

Particle A has mass m and initial velocity v0. A collides with particle B, which has mass M and is initially at rest. After the collision, the particles follow the paths shown in the figure. Calculate φ in degrees, for an ideal elastic collision.
Data: m = 0.7 kg; M = 4.7 kg.

http://imgur.com/UcZciBB

θ=45

Homework Equations

K=\frac{1}{2}m₁v₁²

The Attempt at a Solution

I'm not sure of the most effective route here. I have three equations but I have four unknowns (v₁, v₁', v₂', and \phi)

First I have momentum in the x direction:
m₁v₁=\frac{1}{2}m₁v₁'cos\phi+m₂v₂'cosθ

Second I have momentum in the y direction:
0=m₁v₁'sin\phi-m₂v₂'sin\theta

or

m₁v₁'sin\phi=m₂v₂'sin\theta

Lastly I have the conservation of kinetic energy since this is an elastic collision.

\frac{1}{2}m₁v₁²=\frac{1}{2}m₁v₁'²+\frac{1}{2}m₂v₂'²

I'm not sure of another equation that would help me find reduce this and solve for \phi. Does anyone have any suggestions?

Thank you!

 

[haruspex]

You don't care about the exact velocities, only about the ratios between them. So it will turn out that there are really only three unknowns. Just go ahead and solve the equations.

 

[ehild]

First I have momentum in the x direction:
m₁v₁=\frac{1}{2}m₁v₁'cos\phi+m₂v₂'cosθ

Remove 1/2 from the first momentum equation.

Substitute θ=45°, isolate v₁' or v₂' from the equation for the y components of the momenta, substitute into the other momentum equation, you get both v₁' and v₂' in terms of Φ and v₁. Use these expressions in the energy equation, v1 will cancel.

ehild

 

[oddjobmj]

Remove 1/2 from the first momentum equation.

Substitute θ=45°, isolate v₁' or v₂' from the equation for the y components of the momenta, substitute into the other momentum equation, you get both v₁' and v₂' in terms of Φ and v₁. Use these expressions in the energy equation, v1 will cancel.

ehild

Not sure where that 1/2 came from to begin with. I definitely did not have that written down! Thank you :)

I have tried what you suggested a number of times now both by hand and using wolfram alpha and keep getting \phi=\pin

I made some substitutions to make entering it into wolframalpha easier.

v₁=v
v₁'=z
v₂'=x

mv=mzcos(\phi)+\frac{Mx}{sqrt(2)}

and from the y-component

\frac{Mx}{sqrt(2)}=mzsin(phi)

Take x from the y-component equation and plug it into the x-component equation to solve for z:

z=\frac{Mx}{sqrt(2)msin(phi)}

Do the same for but taking z from the y-component equation and plugging that into the x-component equation to solve for x:

x=\frac{sqrt(2)mvsin(phi)}{Mcos(phi)}

Plugging these x and z values into the energy equation mv²=mz²+Mx²:

mv²=\frac{mv^2}{(cos(phi)+sin(phi))^2}+\frac{2sin^2(phi)}{Mcos^2(phi)}

Plugging in the masses and solving for phi I get p=pi*n

What am I messing up?

 

[ehild]

Not sure where that 1/2 came from to begin with. I definitely did not have that written down! Thank you :)

I have tried what you suggested a number of times now both by hand and using wolfram alpha and keep getting \phi=\pin

that is one solution, which means there is no collision, particle A goes on with its initial velocity and B stays in rest. But there is an other solution.

I made some substitutions to make entering it into wolframalpha easier.

v₁=v
v₁'=z
v₂'=x

mv=mzcos(\phi)+\frac{Mx}{sqrt(2)}

and from the y-component

\frac{Mx}{sqrt(2)}=mzsin(phi)

Take x from the y-component equation and plug it into the x-component equation to solve for z:

z=\frac{Mx}{sqrt(2)msin(phi)}

Do the same for but taking z from the y-component equation and plugging that into the x-component equation to solve for x:

x=\frac{sqrt(2)mvsin(phi)}{Mcos(phi)}

check z and x , I got different formulas. x=\frac{mz\sin(\Phi) \sqrt2}{M}, substitute into the eq. for the x components.
ehild

 

[oddjobmj]

Thank you for your help.

I tried your x originally as well. There are a couple ways to simplify when getting rid of z.

I went through it again and came to the same conclusion on a different path. pi*n

Plugging that x into the x-components equation:

mv=mzcos(phi)+mzsin(phi)

v=z(cos(phi)+sin(phi))

I can use that in the energy equation. I can also plug in my x to the energy equation and it comes out to pi*n again. Fortunately, I plugged the final equation into my calculator instead of wolfram alpha and came up with the correct answer of about 82 degrees.

sin(2*phi)=\frac{1.4}{4.7}*sin^2(phi)

Not sure why alpha doesn't provide the more useful answer.

Thank you very much for your help and patience!

 

[nasu]

Never mind.

 

[ehild]

Thank you for your help.

I tried your x originally as well. There are a couple ways to simplify when getting rid of z.

I went through it again and came to the same conclusion on a different path. pi*n

Plugging that x into the x-components equation:

mv=mzcos(phi)+mzsin(phi)

v=z(cos(phi)+sin(phi))

It is correct now.
z=\frac{v}{\cos(\Phi)+\sin(\Phi)}
Sub it back into the equation for x.
x=\frac{mz\sin(\Phi) \sqrt2}{M}=\frac{mv\sin(\Phi) \sqrt2}{M(\cos(\Phi)+\sin(\Phi))} You can use x and z in the energy equation. m and v² cancel.

(\cos(\Phi)+\sin(\Phi))^2=1+\frac{m}{M} 2 (\sin(\Phi))^2

But (cosΦ +sinΦ)² = 1+2 sinΦ cosΦ.

1 cancels, remains 2 sinΦ cosΦ=2 m/M sin²Φ--> sinΦ(cosΦ-m/MsinΦ)=0
Either sinΦ=0 or m/M sinΦ=cosΦ.

Wolframalpha works for "short" problems. And anyway, it is good for you if you solve the problem yourself.

ehild

 

[bobie]

Data: m = 0.7 kg; M = 4.7 kg.
θ=45

You can simplify you work if you set m=1 (7/7) and V₀ = (M+m) 54/7

θ = 45° makes thing even easier, as px_m becomes 1 (7/7), px _M = py _M = py_m = 47/7 and v_M = √2.

You can find easily: sin\Phi = py_m/ v_m; easier is: cos\Phi =v_m^-1, (√(v_0²- M*v_m²)^-1 = 0.1473... \frac{1}{\sqrt{\frac{2916}{49}- \frac{47*2*7}{7*7}}}= \sqrt\frac{49}{2258}
but you needn't even calculate v_m, extremely simple is:

tan\Phi = \frac{p_y}{p_x} = \frac{47}{7} = 6.7143, → \Phi = 81°. 52885537