Question

Object of mass 3M, moving in the +x direction at speed v0, breaks into two pieces of mass M and 2M, with θ1=46.0º and θ2=31.0º.

Answer 1

1) Determine the x component and the y component of v1 ((that is, of the velocity of the smallest peace, with mass M)) in terms of v0.

2) Determine the x component and the y component of v2 ((that is, of the velocity of the larger peace, with mass 2M)) in terms of v0.

Answer 2

Because momentum is conserved
\(\color{blue}{M~v_{1,i} + 2M~v_{2,i} = M~v_{1,f} + 2M~v_{2,f}} \)

Then, the velocity vector in x and y directions is:
\(\color{blue}{\langle3Mv_o~,~0\rangle~=~\langle M~v_{1,fx}~,~M~v_{1,fy}~\rangle+\langle 2M~v_{2,fx}~,~2M~v_{2,fy}~\rangle} \)

Answer 3

I can cancel out M

\(\color{blue}{v_{1,i} + 2v_{2,i} = v_{1,f} + 2v_{2,f}} \)

\(\color{blue}{\langle3v_o~,~0\rangle~=~\langle v_{1,fx}~,~v_{1,fy}~\rangle+\langle 2v_{2,fx}~,~2v_{2,fy}~\rangle} \)

Answer 4

Is it safe to assume that the angles don't change ?

Answer 5

I would suppose so, because otherwise, how would I solve this?
So I will proceed assuming that the angles don't change, unless anyone explains that they do change ....

Answer 6

Object in motion stays in motion unless acted on by an external force, so you can think of that as saying there's nothing acting on them to change the angle after it breaks apart so it has to keep moving straight ahead.

Answer 7

Oh, thanks! I am usually very bad at physics as regards to knowing the laws and knowing how to apply them(( Alright, I will continue to try to solve it

Answer 8

\(\color{blue}{v_{1,i} + 2v_{2,i} = v_{1,f} + 2v_{2,f}} \)

\(\color{blue}{0=v_{1,fy}+2v_{2,fy}} \)

\(\color{blue}{3v_o=v_{1,fx}+2v_{2,fx}} \)

Answer 9

\(\color{blue}{0=v_{1,f}\sin\theta_1+2v_{2,f}\sin\theta_2}\quad \Rightarrow \quad\color{blue}{\frac{-1}{2}\frac{\sin\theta_1}{\sin\theta_2}v_{1,f}=v_{2,f}} \)

\(\color{blue}{3v_o=v_{1,f}\cos\theta_1+2\frac{-1}{2}\frac{\sin\theta_1}{\sin\theta_2}v_{1,f}\cos\theta_2} \)

\(\color{blue}{3v_o=v_{1,f}\cos\theta_1-\sin\theta_1\tan\theta_2v_{1,f}} \)

\(\color{blue}{3v_o/(\cos\theta_1-\sin\theta_1\tan\theta_2)=v_{1,f}} \)

Answer 10

So, I think I found the first velocity.

Answer 11

\(\color{blue}{3v_o/(\cos\theta_1-\sin\theta_1\tan\theta_2)=v_{1,f}} \)
\(\color{blue}{11.43138~v_o=v_{1,f}} \)

Answer 12

The final velocity x-component of the first peace with mass M,
v_(1,x)=7.94 v_0

but that is incorrect.

Answer 13

These two equations look good, this should be all you need to solve for your two unknowns, the magnitudes of the velocities \(v_1\) and \(v_2\) so let me see if I can find some mistake in your algebra give me a sec

\(\color{blue}{0=v_{1,fy}+2v_{2,fy}} \)

\(\color{blue}{3v_o=v_{1,fx}+2v_{2,fx}} \)

Answer 14

yeah, I am also reworking it...

Answer 15

Also, just curious since you have the answers, can you tell me if these are correct:

\[-5.9699\] and \[4.1690\]

Answer 16

I don't have the answer, I am submitting it, and I got an incorrect answer, but I don't know the answer, and I am also curious to find out what the solution is ...

Answer 17

I will call the first velocity - a, and the second velocity - b, for simplicity.
perhaps my confusion is because I included too many symbols, because I should be decent at algebra as far as I know.

Answer 18

\(\color{blue}{3v_o=a_{f,x}+2b_{f,x}} \)
\(\color{blue}{0=a_{f,y}+2b_{f,y}} \)

\(\color{blue}{3v_o=a_{f}\cos(46)+2b_{f}\cos(31)} \)
\(\color{blue}{0=a_{f}\sin(46)+2b_{f}\sin(31)} \)

from eq 2,
\(\color{blue}{(-1/2)a_{f}\sin(46)/\sin(31)=b_{f}} \)

into eq 1,
\(\color{blue}{3v_o=a_{f}\cos(46)+2(-1/2)a_{f}[\sin(46)/\sin(31)]\cos(31)} \)
\(\color{blue}{3v_o=a_{f}\cos(46)+2(-1/2)a_{f}\sin(46)\cot(31)} \)

first error, cot instead of tan

Answer 19

\(\color{blue}{3v_o=a_{f}[\cos(46)-\sin(46)\cot(31)]} \)
\(\color{blue}{v_o\times 3/[\cos(46)-\sin(46)\cot(31)]=a_{f}} \)

\(\color{blue}{-5.96986~v_o=a_{f}} \)

the first velocity

Answer 20

Well looks like we're consistently wrong then haha

Answer 21

Well, assuming we're not right that is... heh

Answer 22

the second velocity

\(\color{blue}{0=-5.96986~v_o\sin(46)+2b_{f}\sin(31)}\)
\(\color{blue}{5.96986~v_o\sin(46)=2b_{f}\sin(31)}\)
\(\color{blue}{5.96986~v_o\sin(46)/[2\sin(31)]=b_{f}}\)
\(\color{blue}{v_o~4.16897=b_{f}}\)

Answer 23

Those are the velocities, if they are correctly, which I see no reason they aren't now that I carefully gone over this.

Answer 24

\(\color{blue}{4.16897~v_o=b_{f}}\)
\(\color{blue}{-5.96986~v_o=a_{f}}\)

and then the components are as follows:

Answer 25

\(\color{blue}{-5.96986\cos(46)~v_o=a_{f_x} }\)
\(\color{blue}{-4.147~v_o=a_{f_x} }\)

Answer 26

For the sake of zeus, how am I incorrect again ????

Answer 27

http://i.imgur.com/DS99qAU.jpg

sorry I forgot the picture

Answer 28

I will restart once again((

Answer 29

\(\color{blue}{3v_o=a_{f,x}+2b_{f,x}}\)
\(\color{blue}{0=a_{f,y}+2b_{f,y}}\)

\(\color{blue}{3v_o=(\cos\theta_1)a_{f}+(2\cos\theta_2)b_{f}}\)
\(\color{blue}{0=(-\sin\theta_1)a_{f}+(2\sin\theta_2)b_{f}}\)

Answer 30

I think I found another error, and hopefully the last one.

Answer 31

I guess they should both be positive if they're magnitudes, maybe that's it?

Answer 32

Ok, so continuing from there, I have,
\(\color{blue}{(\sin\theta_1)a_{f}=(2\sin\theta_2)b_{f}}\)
\(\color{blue}{a_{f}=(2\sin\theta_2)/(\sin\theta_1)~b_{f}}\)

\(\color{blue}{3v_o=(\cos\theta_1)a_{f}+(2\cos\theta_2)b_{f}}\)
\(\color{blue}{3v_o=(\cos\theta_1)(2\sin\theta_2)/(\sin\theta_1)~b_{f}+(2\cos\theta_2)b_{f}}\)
\(\color{blue}{3v_o=\{(\cot\theta_1)(2\sin\theta_2)+2\cos\theta_2\}~b_{f}}\)

\(\color{blue}{1.1073921~v_o=~b_{f}}\)

Answer 33

yes, the gods have taken my prayer!

0.9492 v_0 for the x-comp of second velocity!

Answer 34

again,

0.57034 v_0 for y-comp of second velocity.

Answer 35

finished !