I have a conservation of momentum equation Q

[Divergent13]

I have a conservation of momentum equation Q!

A 0.25kg skeet (clay target) is fired at an angle of 30 degrees to the horizon with a speed of 30 m/s. When it reaches its maximum height, it is hit from below by a 15g pellet traveling vertically upward at a speed 200m/s, the pellet is imbedded in the skeet.

So they want to know how much higher the skeet would go, and how much EXTRA horizontal distance the skeet would travel because of that collision...

I believe you can get the height from using conservation of energy. Then you can use what you know about projectiile motion to find all the different time intervals involved. From there, Distance in the x direction is simply Vx*t

I am just not sure if I am getting the correct numbers? What would you guys do?

 

[arildno]

Step 1.
Energy conservation up to maximal height h for the skeet:
\frac{1}{2}V_{0}^{2}=\frac{1}{2}V_{0,x}^{2}+gh
(V_{0} is initial velocity, with components V_{0,x},V_{0,y})
Step 2. Conservation of momentum in inelastic collision.
This will yield a non-zero upwards velocity component.
Step 3. Book-keeping:
Record horizontal distance already traveled, X_{1}, and vertical and horizontal velocities after collision.
Step 4.
Solve for landing position as a function of time, using parameters obtained in Step 3.

 

[Theelectricchild]

But since Vo x is constant why would you use that equation to solve it?

 

[arildno]

Because I forgot to eliminate it on both sides of the equation..

 

[Divergent13]

I get an impact height of 11.43 m. Is this correct? (Using your equation.)

 

[arildno]

If 11.43=\frac{(30\sin30)^{2}}{2g}, then it is correct, as long as you with "impact height" means the height of the sleet when the proctile hits it.

 

[Divergent13]

Cool then I just use conservation of momentum to find the new Vx and Vy and treat it like an ordinary 2D kinematics problem.

 

[arildno]

Yes; that would be it.