The problem is as follows:
A body of mass $m$ is moving with a speed of $v$ makes an elastic collision with another which was initially at rest and as a result it gets expelled by the object which was at rest. Then the body of mass $m$ makes an angle of $90^{\circ}$ to the initial direction of its motion and its speed is reduced to $\frac{v}{2}$. Find the mass of the second body.
The alternatives are as follows:
$\begin{array}{ll} 1.&\frac{5m}{3}\\ 2.&\frac{5m}{4}\\ 3.&\frac{5m}{2}\\ 4.&\frac{7m}{3}\\ 5.&4m\\ \end{array}$
This particular problem doesn't offer a sketch or drawing so I made one according to how I did saw the situation:
The figure from below illustrates the two events separately. The first occurs before the collision and the second pictured in the bottom is the event after the collision.
For this situation I'm assuming that the object whose mass is unknown must be bigger in its mass to deflect the other particle and send it straight upwards, while the other will move with some deflection from its original position with an angle which I labeled as $\phi$.
The rest it is just my understanding of the application of the conservation of momentum for each axis as follows:
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:
$pi=pf$
$mv=m(0)+Mu_{2}\cos\phi$
$u_{2}=\frac{mv}{Mcos\phi}$
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
$\frac{3mv^2}{4}=Mu_{2}^2$
$u_{2}^{2}=\left(\frac{mv}{2M\frac{2}{\sqrt{5}}}\right)^2$
$u_{2}^{2}=\frac{5m^{2}v^{2}}{16M^{2}}$
Then plugging into the earlier expression:
$\frac{3mv^{2}}{4}=\frac{5m^{2}v^{2}}{16M}$
Then:
$M=\frac{5m}{12}$
But if I were to use the cosine, I got to this:
$u_{2}^{2}=\left(\frac{mv}{M\left(\frac{1}{\sqrt{5}}\right)}\right)^2$
$\frac{3mv^{2}}{4}=\frac{5m^{2}v^{2}}{M}$
$M=\frac{20m}{3}$
Then I obtain two answers!. Why is this happening?
Can somebody help me with my method at pointing exactly where did I derailed from the logic or what did I overlooked?. I would like to understand exactly if I did understood the physical part right and if the mathematical operations are alright. Can someone help me with this as well?. I appreciate that a solution to accompany the answer using my approach.