Finding Angle and Intensity of Split Rocket at Maximum Height

[doktorwho]

Homework Statement

A rocket is fired from from the ground at initial velocity of $v_0$ and at an angle $\theta$. At its highest height it splits into 2 parts of equal masses. The first part is fired straight up and at velocity $v_0/2$. Find the angle and intensity of the second part.

Homework Equations

3. The Attempt at a Solution [/B]
At its maximum height the velocity has only the x component which is $v_0cos\theta$. The momentum is $p_0=mv_0cos\theta$. Since it has only the x component and the part which divides has the y component the second part must have an equal y component. Mening the $v_{2y}=v_{1y}=v_0/2$. That would make the change in the y component of the momentum $0$. Since that starting condition has total momentum in x direction the $v_2$ part must have the x component equal to the $v_0cos\theta$. The intensity would then be $\sqrt{v_{2y}^2+v_{2x}^2}$. Does this seem right couse when i use it in the problem given by the books where this intial velocity is $1000$ the angle $60$ i get $v_y=500\sqrt{2}$ and the book gives $v_y=1118$?

 

[TomHart]

When the problem says that the first part is fired straight up, I think it means straight up relative to someone on the ground - not straight up relative to the moving projectile.

 

[Merlin3189]

Your method sounds correct in general, though I get a different answer from you when using that approach. I think the book is correct.
Perhaps you could show your working in detail.
I think you may be going wrong in the horizontal component, what you label v_2y

 

[doktorwho]

Your method sounds correct in general, though I get a different answer from you when using that approach. I think the book is correct.
Perhaps you could show your working in detail.
I think you may be going wrong in the horizontal component, what you label v_2y

At the max heigth the velocity has only the x-component which is $v_0cos\theta=500m/s$. Since it then splits into two parts they both must have 0 net y velocity and 500m/s in the x-component. $v_{1y}=500m/s$ and $v_{1x}=0$. From there it follows that the vertical component of the second velocity must be $500m/s$ in order for the y-components to cancel out.
The x component must be equal to $m*500m/s$. So $\frac{m}{2}v_{2x}=m500$. So it should be $1000m/s$ Now i get it right. I forgot the fact that the mass splits so i thought that i have full mass of velocity 2. Good now.

 

[Merlin3189]

Exactly where I went wrong myself at first attempt!