How do I find the mass of a body when two bodies collide and one makes an angle?

Jun 2017
302
6
Lima, Peru
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:

$\begin{array}{ll}
1.&\frac{5m}{3}\\
2.&\frac{5m}{4}\\
3.&\frac{5m}{2}\\
4.&\frac{7m}{3}\\
5.&4m\\
\end{array}$

Can somebody help me here, what exactly should I do to relate the angle which is given to find the mass of the body?

Since it mentions that it is an elastic collision, I think that it is telling me that I should preserve the kinetic energy, but this complicated the calculations. How can I solve this problem?
 
Last edited by a moderator:

topsquark

Math Team
May 2013
2,385
996
The Astral plane
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
 
Jun 2017
302
6
Lima, Peru
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
It would help me better to have a guide on where to put those components and exactly when should be considered the angle. This topic I am struggling with. :(
 

topsquark

Math Team
May 2013
2,385
996
The Astral plane
Here's the sketch. I'm calling the blue ball your mass m, and the one initially at rest is red. I'm going to use the "usual" coordinate system (+x to the right and +y upward.) Above the line is the before picture and below it is after the collision.

Conservation of momentum:
x direction:
\(\displaystyle P_{ix} = P_{fx} \implies mv + (0) = (0) + m_2 v_{2x}\)

y direction:
\(\displaystyle P_{iy} = P_{fy} \implies (0) + (0) = m \left ( \frac{v}{2} \right ) + m_2 v_{2y}\)

Conservation of Kinetic energy:
\(\displaystyle KE_{i1} + KE_{i2} = KE_{f1} + KE_{f2} \implies \frac{1}{2} m v^2 + (0) = \frac{1}{2} m \left ( \frac{v}{2} \right ) ^2 + \frac{1}{2} m_2 v_2 ^2\)

Note that we have components for the final velocity of \(\displaystyle m_2\): \(\displaystyle v_{2x} = v_2 ~ \cos( \theta )\) and \(\displaystyle v_{2y} = -v_2 ~ \sin( \theta )\). This means we have the system of equations:
\(\displaystyle mv = m_2 v_2 ~ \cos( \theta )\)
\(\displaystyle 0 = \frac{1}{2} mv - m_2 v_2 ~ \sin( \theta )\)
\(\displaystyle \frac{1}{2} m v^2 = \frac{1}{8} m v^2 + \frac{1}{2} m_2 v_2 ^2\)

This is going to be a bit of a bear for you to solve, as I mentioned earlier. Let's make sure we agree on the equations first.

Mind you, this kind of diagram is going to occur a lot in your practice problems. What I did above is pretty general so please make sure you understand it.

-Dan
collison.jpg
 
Jun 2017
302
6
Lima, Peru
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. :rolleyes:
 
Jun 2017
302
6
Lima, Peru
Here's the sketch. I'm calling the blue ball your mass m, and the one initially at rest is red. I'm going to use the "usual" coordinate system (+x to the right and +y upward.) Above the line is the before picture and below it is after the collision.

Conservation of momentum:
x direction:
\(\displaystyle P_{ix} = P_{fx} \implies mv + (0) = (0) + m_2 v_{2x}\)

y direction:
\(\displaystyle P_{iy} = P_{fy} \implies (0) + (0) = m \left ( \dfrac{v}{2} \right ) + m_2 v_{2y}\)

Conservation of Kinetic energy:
\(\displaystyle KE_{i1} + KE_{i2} = KE_{f1} + KE_{f2} \implies \dfrac{1}{2} m v^2 + (0) = \dfrac{1}{2} m \left ( \dfrac{v}{2} \right ) ^2 + \dfrac{1}{2} m_2 v_2 ^2\)

Note that we have components for the final velocity of \(\displaystyle m_2\): \(\displaystyle v_{2x} = v_2 ~ cos( \theta )\) and \(\displaystyle v_{2y} = -v_2 ~ sin( \theta )\). This means we have the system of equations:
\(\displaystyle mv = m_2 v_2 ~ cos( \theta )\)
\(\displaystyle 0 = \dfrac{1}{2} mv - m_2 v_2 ~ sin( \theta )\)
\(\displaystyle \dfrac{1}{2} m v^2 = \dfrac{1}{8} m v^2 + \dfrac{1}{2} m_2 v_2 ^2\)

This is going to be a bit of a bear for you to solve, as I mentioned earlier. Let's make sure we agree on the equations first.

Mind you, this kind of diagram is going to occur a lot in your practice problems. What I did above is pretty general so please make sure you understand it.

-Dan
View attachment 10736
Sorry I do not have much time right now to comment about your answer. I was hoping to show my effort first. If you happen to have time in the evening I'll be posting an update on my earlier comment. :)
 

skeeter

Math Team
Jul 2011
3,212
1,734
Texas
using topsquark's momentum equations ...

$mv = m_2v_{2x} \implies v_{2x} = \dfrac{mv}{m_2}$

$m \cdot \dfrac{v}{2} = -m_2v_{2y} \implies v_{2y} = -\dfrac{mv}{2m_2}$

KE is conserved ...

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}m\left(\dfrac{v}{2}\right)^2 + \dfrac{1}{2}m_2\left(v_{2x}^2 + v_{2y}^2\right)$

substituting for the two components of $v_2$ and doing the grunt work algebra yields the equation ...

$3 = \dfrac{5m}{m_2} \implies m_2 = \dfrac{5m}{3}$
 
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Jun 2017
302
6
Lima, Peru
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. Give it a shot and show us what you get.

-Dan
As you suggested I did that and as promised in an earlier message here's how I made the interpretation:



The equations I posted earlier follow the logic from the sketch I made. Sorry there was a delay into that. But I had to make a drawing after returning from my classes. :)
 
Jun 2017
302
6
Lima, Peru
using topsquark's momentum equations ...

$mv = m_2v_{2x} \implies v_{2x} = \dfrac{mv}{m_2}$

$m \cdot \dfrac{v}{2} = -m_2v_{2y} \implies v_{2y} = -\dfrac{mv}{2m_2}$

KE is conserved ...

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}m\left(\dfrac{v}{2}\right)^2 + \dfrac{1}{2}m_2\left(v_{2x}^2 + v_{2y}^2\right)$

substituting for the two components of $v_2$ and doing the grunt work algebra yields the equation ...

$3 = \dfrac{5m}{m_2} \implies m_2 = \dfrac{5m}{3}$
I am just wondering, can you take a look into my work?. In your analysis you did not used any sort of angle as I did. Wouldn't make sense that the same answer could be obtained if we would use angles?.