Elastic collision with pendulum

[Jrlinton]

Homework Statement

A steel ball of mass 0.890 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal, as shown in the figure. At the bottom of its path, the ball strikes a 2.50 kg steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.

Homework Equations

The Attempt at a Solution

So first found the velocity of the ball just before it collided with the block:
V=(2*g*L*(1-cosΘ))^0.5
V=(2*9.81*0.5*(1-cos45))^0.5
V=2.873 m/s

So then I used that to find the velocity of the block using elastic collision formulas:
V2f= (2*m1)/(m1+m2)*V1i
= (2*0.89)/(0.89+2.5)*2.873
=1.51 m/s

And then for the velocity of the ball following the collision:
V1f=(m1-m2)/(m1+m2)*V1i
=(0.89-2.5)/(0.89+2.5) * 2.873
=-1.37 m/s

I used these solutions (the absolute value of the vel of the ball as it asked for speed) and that was incorrect.
Now I did try and double check myself and found that with these numbers momentum is conserved so what am I missing here?

 

[CWatters]

I haven't checked your attempt at a solution in any detail but a Cos(45) term seems odd given the rope starts horizontal and the collision is at the bottom?

You mention a figure but none provided.

 

[jbriggs444]

The ball is then released when the cord is horizontal, as shown in the figure.
[...]
V=(2*g*L*(1-cosΘ))^0.5
V=(2*9.81*0.5*(1-cos45))^0.5

Without seeing the drawing, it is hard to be sure, but I see no 45 degree angle in the problem description.

Edit: Drat you, speedy @CWatters!

 

[Jrlinton]

Youre right. it should've been 90 degrees as it was raised to the horizontal. I am not sure why I used 45. Other than that the method should hold, no?

 

[Jrlinton]

So using the correct Θ of 90 degrees I get the final velocity of the block to be 1.64 m/s and that of the ball to be -1.49 m/s.