2 derivations using the Conservation of Linear Momentum

[auk411]

Homework Statement

I have 2 equations. I need to derive two different equations from them.

Derive (1): v₁₀²= v_1f² + v_f2²+ 2v_1fv_2fcos(φ+θ)

(2): v_2f = v₁₀sin[tex]\theta[/tex]/sin(theta - phi)

Homework Equations

v₁₀= v_1fcosθ + v_2fcosφ
0 = v_2fsinφ - v_2fsinθ

The Attempt at a Solution

I know for one you are going to have to square both sides of both equations. I figure I'm going to have to use some trig identities (actually for both), but I can't seem to find anywhere to go.

These are part of larger problems that I have worked through and set up but I can't seem to do the algebra/trig.

 

[cepheid]

Homework Equations

v₁₀= v_1fcosθ + v_2fcosφ
0 = v_2fsinφ - v_2fsinθ

It look like you're considering the case of an object that is traveling entirely in the horizontal direction when it suddenly "explodes" into two pieces, one of which moves off at an angle of theta above the horizontal, and the other of which travels off at an angle of phi below the horizontal. Am I right? The relevant equations you have posted then come from conservation of momentum in the horizontal and vertical directions respectively. Correct? (EDIT: Hmm...but what about the masses? )

EDIT 2: No, I figured it out. It wasn't an explosion. It was a collision. Mass 1 one traveling horizontally when it collided with *stationary* object 2 of EQUAL mass, which is why the masses cancel from both sides. They then traveled off obliquely, right?

Also, should the '2' in red actually be a '1'?

 

[cepheid]

I know for one you are going to have to square both sides of both equations. I figure I'm going to have to use some trig identities (actually for both), but I can't seem to find anywhere to go.

Yeah. Have you done that (squared both sides?) What did you get? Can you see how might manipulate that to turn it into the result?

Here's another hint: work backwards from the result. It has cos(φ+θ) in it. There are some trigonometric identities known as the "sum and difference formulae" that tell you what the sine and cosine of a sum or difference of two angles is. Applying that to the result, what do you get?

 

[auk411]

Yeah. Have you done that (squared both sides?) What did you get? Can you see how might manipulate that to turn it into the result?

Here's another hint: work backwards from the result. It has cos(φ+θ) in it. There are some trigonometric identities known as the "sum and difference formulae" that tell you what the sine and cosine of a sum or difference of two angles is. Applying that to the result, what do you get?

Thanks, I got it.

 

[rwisz]

I can provide a response to your question by using the conservation of linear momentum, which states that the total linear momentum of a system remains constant unless acted upon by an external force. In other words, the initial momentum of a system must equal the final momentum.

Derivation (1):
Starting with the given equations:
v10= v1fcosθ + v2fcosφ
0 = v2fsinφ - v2fsinθ

We can square both sides of the first equation to get:
v102= (v1fcosθ)^2 + 2v1fcosθv2fcosφ + (v2fcosφ)^2

Using the trigonometric identity cos(A+B) = cosAcosB - sinAsinB, we can rewrite the second term as:
2v1fcosθv2fcosφ = 2v1fv2fcos(φ+θ)

Substituting this into our equation, we get:
v102= (v1fcosθ)^2 + 2v1fv2fcos(φ+θ) + (v2fcosφ)^2

Now, we can use the second equation, 0 = v2fsinφ - v2fsinθ, to substitute for v2fcosφ and v2fsinφ:
v102= (v1fcosθ)^2 + 2v1fv2fcos(φ+θ) + (v2fsinθ)^2 - (v2fsinθ)^2

Using the trigonometric identity sin^2A + cos^2A = 1, we can simplify the equation to get:
v102= (v1fcosθ)^2 + (v2fsinθ)^2 + 2v1fv2fcos(φ+θ) - (v2fsinθ)^2

Finally, we can rearrange the terms to get the desired equation:
v102= v1f2 + vf22 + 2v1fv2fcos(φ+θ)

Derivation (2):
Starting with the given equation:
v2f = v10sin\theta/sin(theta - phi)

We can rewrite the equation as:
v2f = v10sin\theta/sin\thetacos\phi - cos\theta sin\phi

Using the trigonometric identity