Conservation of momentum in two dimensions

[keroberous]

Homework Statement

A bomb initially at rest is exploded into three pieces on a smooth, horizontal surface. Two pieces fly off at a 60° angle to each other, a 2.0 kg piece at 20 m/s and a 3.0 kg piece at 12 m/s. The third piece flies off at 30 m/s with an unknown direction. Determine the direction and mass of the third piece.

Homework Equations

conservation of linear momentum
$m_1v_{1x}+m_2v_{2x}+m_3v_{3x}=m_1v'_{1x}+m_2v'_{2x}+m_3v'_{3x}$
and
$m_1v_{1y}+m_2v_{2y}+m_3v_{3y}=m_1v'_{1y}+m_2v'_{2y}+m_3v'_{3y}$

3. The Attempt at a Solution

I have a full solution, but the question doesn't give an answer so I'm not sure if I'm correct.
I have attached a picture of how I set up the axes for reference.

Since the bomb was initially at rest my equations become:
$0=m_1v'_{1x}+m_2v'_{2x}+m_3v'_{3x}\\0=(2.0)(20)+(3.0)(12\cos(60))+m_3(-30)\cos\theta\\0=58-30m_3\cos\theta\\m_3=\frac {58} {30\cos\theta}$
and

$0=m_1v'_{1y}+m_2v'_{2y}+m_3v'_{3y}\\0=(2.0)(0)+(3.0)(12\sin(60))+m_3(-30)\sin\theta\\0=31.177-30m_3\sin\theta\\m_3=\frac {31.177} {30\sin\theta}$

Now equating and solving for theta:

$\frac {58} {30\cos\theta}=\frac {31.177} {30\sin\theta}\\\tan\theta=\frac{31.177}{58}\\\theta=28$

And finally, solving for the mass:

$m_3=\frac {58} {30\cos\theta}\\m_3=\frac {58} {30\cos(28)}\\m_3=2.2$

I'd appreciate it if you could give insight into my line of reasoning and final solutions. Thanks!

 

[Doc Al]

I didn't check your arithmetic, but your approach and equations look good to me.

 

[Ray Vickson]

Homework Statement

A bomb initially at rest is exploded into three pieces on a smooth, horizontal surface. Two pieces fly off at a 60° angle to each other, a 2.0 kg piece at 20 m/s and a 3.0 kg piece at 12 m/s. The third piece flies off at 30 m/s with an unknown direction. Determine the direction and mass of the third piece.

Homework Equations

conservation of linear momentum
$m_1v_{1x}+m_2v_{2x}+m_3v_{3x}=m_1v'_{1x}+m_2v'_{2x}+m_3v'_{3x}$
and
$m_1v_{1y}+m_2v_{2y}+m_3v_{3y}=m_1v'_{1y}+m_2v'_{2y}+m_3v'_{3y}$

3. The Attempt at a Solution

I have a full solution, but the question doesn't give an answer so I'm not sure if I'm correct.
I have attached a picture of how I set up the axes for reference.

Since the bomb was initially at rest my equations become:
$0=m_1v'_{1x}+m_2v'_{2x}+m_3v'_{3x}\\0=(2.0)(20)+(3.0)(12\cos(60))+m_3(-30)\cos\theta\\0=58-30m_3\cos\theta\\m_3=\frac {58} {30\cos\theta}$
and

$0=m_1v'_{1y}+m_2v'_{2y}+m_3v'_{3y}\\0=(2.0)(0)+(3.0)(12\sin(60))+m_3(-30)\sin\theta\\0=31.177-30m_3\sin\theta\\m_3=\frac {31.177} {30\sin\theta}$

Now equating and solving for theta:

$\frac {58} {30\cos\theta}=\frac {31.177} {30\sin\theta}\\\tan\theta=\frac{31.177}{58}\\\theta=28$

And finally, solving for the mass:

$m_3=\frac {58} {30\cos\theta}\\m_3=\frac {58} {30\cos(28)}\\m_3=2.2$

I'd appreciate it if you could give insight into my line of reasoning and final solutions. Thanks!

The method is OK, but a bit long-winded. Your answers are not accurate because of premature rounding: the answer is $\theta = 28.259^o,$ not $28^o.$ That changes the sine and cosine a bit, so you get a slightly incorrect $m_3$. The answer should be $m_3 = 2.1949.$ Of course, when you round off your answers to 2 significant figures you do have $\theta \approx 28^o$ and $m_3 \approx 2.2,$ just as you said, but it is always recommended that you keep more accuracy during you calculations, and just round at the end. (Of course, it may also be the case that you kept more figures during the working but just typed out 2 sig-fig results in your writeup, but I cannot tell that

Homework Statement

A bomb initially at rest is exploded into three pieces on a smooth, horizontal surface. Two pieces fly off at a 60° angle to each other, a 2.0 kg piece at 20 m/s and a 3.0 kg piece at 12 m/s. The third piece flies off at 30 m/s with an unknown direction. Determine the direction and mass of the third piece.

Homework Equations

conservation of linear momentum
$m_1v_{1x}+m_2v_{2x}+m_3v_{3x}=m_1v'_{1x}+m_2v'_{2x}+m_3v'_{3x}$
and
$m_1v_{1y}+m_2v_{2y}+m_3v_{3y}=m_1v'_{1y}+m_2v'_{2y}+m_3v'_{3y}$

3. The Attempt at a Solution

I have a full solution, but the question doesn't give an answer so I'm not sure if I'm correct.
I have attached a picture of how I set up the axes for reference.

Since the bomb was initially at rest my equations become:
$0=m_1v'_{1x}+m_2v'_{2x}+m_3v'_{3x}\\0=(2.0)(20)+(3.0)(12\cos(60))+m_3(-30)\cos\theta\\0=58-30m_3\cos\theta\\m_3=\frac {58} {30\cos\theta}$
and

$0=m_1v'_{1y}+m_2v'_{2y}+m_3v'_{3y}\\0=(2.0)(0)+(3.0)(12\sin(60))+m_3(-30)\sin\theta\\0=31.177-30m_3\sin\theta\\m_3=\frac {31.177} {30\sin\theta}$

Now equating and solving for theta:

$\frac {58} {30\cos\theta}=\frac {31.177} {30\sin\theta}\\\tan\theta=\frac{31.177}{58}\\\theta=28$

And finally, solving for the mass:\pi

I'd appreciate it if you could give insight into my line of reasoning and final solutions. Thanks!

Your answer is OK and you have used a correct method.

 

[Orodruin]

So let me just offer an alternative solution that I find more elegant and coordinate independent. Instead of working with $m_3$ and an angle, first consider that the momentum of 3 is $m_3 \vec v_3$. Conservation of momentum gives
$$
m_3\vec v_3 = -m_1\vec v_1 -m_2\vec v_2.
$$
Squaring this relation directly gives $m_3$. The direction is then easily found through the sine theorem as the three momenta form a triangle.

 

[keroberous]

Thanks all for your help!