Some general advice. First sketch a before and after picture, labeling a +x and +y coordinate system. Then you have both conservation of momentum and conservation of energy to work with. Get your velocity components from the diagrams.

No, it's probably not going to be terribly easy, but the Math isn't that bad here, it's just fussy with the details. I'd start by putting the information into the momentum equations and use those in the kinetic energy equations.

Give it a shot and show us what you get.

-Dan

@topsquark If you happen to have time to check this...

I did not have enough time to offer a before and after sketch but here's how I made the interpretation:

First there's a conservation of momentum:

$v$ and $m$ are given:

Now I'm letting $M$ to be the mass of the bigger object which the first one collided to and then gets expelled.

There will be two components $x$ and $y$:

For the first component $\textrm{x-axis}$ it will be like this:

$p_{i}=p_{f}$

$mv=m(0)+Mu_2\cos \phi$

$u_{2}=\frac{mv}{M\cos\phi}$

For the second component $\textrm{y-axis}$ it will be like this:

$0=m\frac{v}{2}+Mu_{2}\sin\phi$

$u_2=-\frac{mv}{2M\sin\phi}$

This should be understood as $u_{2}$ the final speed of the bigger object:

Then there will be the conservation of mechanical energy but before that... let's wait. I can find the angle from those two expressions:

$-\frac{mv}{2M\sin\phi}=\frac{mv}{M\cos\phi}$

which translates into:

$\tan\phi=-\frac{1}{2}$

But for the tangent to be negative the angle will lie in the second of the fourth quadrant. Which makes sense to choose the fourth quadrant as it will be in the direction where the second object would be heading to after the collision. (Please correct to me this statement if it is not right.)

From this it can be inferred that I can use any of those two expressions to get the values of the trigonometric functions:

$\cos\phi=+\frac{1}{\sqrt{5}}$

$\sin\phi=-\frac{2}{\sqrt{5}}$

These signs must be swapped in order to satisfy the value of the tangent, for the fourth quadrant I'm letting the cosine positive and the sine negative. But since any of the two will be squared in the end, it won't matter.

Then we're all set to move to the next phase:

The conservation of mechanical energy:

Here all are scalars so we do not worry about the direction.

This will be as follows:

$E_{i}=E_{f}$

$\frac{1}{2}mv^2=\frac{1}{2}m\left(\frac{1}{2}v\right)^2+\frac{1}{2}Mu_{2}^2$

$\frac{3mv^2}{8}=\frac{1}{2}Mu_{2}^2$

We're getting closer as two alternatives have the number three in the denominator:

Now all that is left is to plug in the value of $u_{2}$:

Here's the part where I'm stuck with:

Let's suppose I'm square the $u_2$ from sine function

$\frac{3mv^2}{4}=Mu_{2}^2$

$u_{2}^2=\left(\frac{mv}{2M\left(\frac{2}{\sqrt{5}}\right)} \right)^2$

$u_{2}^2=\frac{5m^2v^2}{16M^2}$

Then plugging into the earlier expression:

$\frac{3mv^2}{4}=\frac{5m^2v^2}{16M}$

Then:

$M=\frac{5m}{12}$

But if I were to use the cosine, I got to this:

$u_{2}^2=\left(\frac{mv}{M\left(\frac{1}{\sqrt{5}}\right)} \right)^2$

$\frac{3mv^2}{4}=\frac{5m^2v^2}{1M}$

$M=\frac{20m}{3}$

Then I obtain two answers!. Why is this happening?

Right now is 4:47pm in my time zone and I'm heading off to my class, so by the evening I'll add my understanding in a drawing of how I'm making the interpretation of this problem. Sorry I cannot offer more. I'd appreciate if

@skeeter can also take a look into my effort and find where did I got lost?.

Gee, I never thought that this particular topic of my course in physics be like this.