Conservation of Momentum of a bomb shell

[SammyS]

p0=2m*v0
p1=m*v0

Don't forget they have directions associated with them.

Now, find p₂ ?

 

[heycoa]

so the direction of p0 is in the x-direction, and p1 is in the y-direction?

 

[SammyS]

so the direction of p0 is in the x-direction, and p1 is in the y-direction?

Yes ...

 

[heycoa]

well p2=v2*m

and when i solve for v2 i get 2*v0(x hat) + v0(y hat)

 

[SammyS]

well p2=v2*m

and when i solve for v2 i get 2*v0(x hat) + v0(y hat)

From that I get $\ \vec{p_2}=2\,m\,v_0\hat{x}+m\,v_0\hat{y}\ .$

Adding p₁ and p₂ should give p₀, right?

 

[heycoa]

Then it should be minus the second term (-m*v0*yhat), right?

 

[SammyS]

Then it should be minus the second term (-m*v0*yhat), right?

Yes, for p₂.

What does this give you for the vector, v₂, and its direction, and its magnitude, v₂ ?

 

[heycoa]

for v2 i get 2*v0(x hat) + v0(y hat)

i calculated the magnitude to be v2=v0*sqrt(5)

does this appear to be correct?

 

[SammyS]

for v2 i get 2*v0(x hat) + v0(y hat)

i calculated the magnitude to be v2=v0*sqrt(5)

does this appear to be correct?

Yes !

 

[heycoa]

Ok excellent!

So I apparently need to read these questions more carefully and define my coordinate systems.

I can't thank you enough for taking the time and having the patience to work with and follow up with me. Thank you very much

 

[SammyS]

Ok excellent!

So I apparently need to read these questions more carefully and define my coordinate systems.

I can't thank you enough for taking the time and having the patience to work with and follow up with me. Thank you very much

You're welcome.

I hope I wasn't being too difficult at times.

 

[heycoa]

no you're great! :)