Conservation of momentum in an oblique launch and projectile explosion

AI Thread Summary
The discussion revolves around the conservation of momentum in a projectile explosion, focusing on the calculations of velocity components and their implications. Participants analyze the equations derived from the conservation of momentum, particularly emphasizing the horizontal and vertical components of velocity. There is a debate about the treatment of the x-component of momentum, with concerns raised about assuming it remains unchanged post-explosion. Clarifications are made regarding the use of angles and the center of horizontal momentum frame to simplify calculations. Overall, the conversation highlights the complexities of applying momentum conservation principles in this scenario.
TheGreatDeadOne
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Homework Statement
A projectile of mass m is fired with initial velocity of module v_0 at an elevation angle of 45◦. The projectile explodes in the air in two pieces of masses m/3 and 2m/3. The pieces continue to move in the same plane as the entire projectile and reach the ground together. The smaller piece falls at a distance of 3(v_0)^2 /2g from the launch point. Determine the range of the largest chunk. Neglect air resistance
Relevant Equations
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This problem I already solved using another resource (just get the coordinate of the center of mass reach and from it, get to the larger mass. R = (3v02) / (4g)). But I'm having some trouble calculating using moment conservation. Here what I've done so far:

$$ 3\vec v_0 = \vec v_1 +2\vec v_2 $$

As the fragments fall in the same time interval, the vertical components of their velocities are the same, since in the act of the explosion, they depart from the same height:
$$ 3(\cos{\theta},\sin{\theta})= (v_{1,x},v_{1,y}) + 2(v_{2,x},v_{2,y})
=(v_{1,x},v_{1,y}) + 2(v_{2,x},v_{1,y})$$
$$ \therefore v_0 \sin{\theta}=v_{1,y}$$
In addition, we can equalize the fall time intervals of each fragment using the horizontal component of each fragment (uniform movement):
$$ (v_{1,x},v_{1,y})= (v_0\cos{\theta},v_{0}\sin{\theta})$$
$$ (v_{2,x},v_{2,y})= (2v_0\cos{\theta},2v_{0}\sin{\theta})$$
Range:
$$t_1=t_2$$
$$\frac{R_1}{v_{1,x}}=\frac{R_2}{v_{2,x}}$$
$$R_2=\frac{R_1 v_{2,x}}{v_{1,x}}=\frac{3v_0^2}{g}$$
 
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I don't understand what you're doing. You could use the centre of horizontal momentum frame. Also, the y-momentum can be taken out of the equations, because of the equal time condition, leaving you to focus on the x-momentum.
 
PeroK said:
I don't understand what you're doing. You could use the centre of horizontal momentum frame. Also, the y-momentum can be taken out of the equations, because of the equal time condition, leaving you to focus on the x-momentum.
I used the center of horizontal momentum frame. I just used the y-moment to find the angle of the velocity component, so I can write the velocity in the horizontal direction. I'll edit the equations to make it clearer
 
TheGreatDeadOne said:
I used the center of horizontal momentum frame. I just used the y-moment to find the angle of the velocity component, so I can write the velocity in the horizontal direction. I'll edit the equations to make it clearer
I think if you are given ##\theta = \frac \pi 4##, then in a problem like this you should be using ##\cos \theta = \frac 1 {\sqrt{2}}##.
 
TheGreatDeadOne said:
$$ 3(\cos{\theta},\sin{\theta})= (v_{1,x},v_{1,y}) + 2(v_{2,x},v_{2,y})
=(v_{1,x},v_{1,y}) + 2(v_{2,x},v_{1,y})$$
$$ \therefore v_0 \sin{\theta}=v_{1,y}$$
In addition, we can equalize the fall time intervals of each fragment using the horizontal component of each fragment (uniform movement):
$$ (v_{1,x},v_{1,y})= (v_0\cos{\theta},v_{0}\sin{\theta})$$

If I have correctly understood what you are doing, I think your key mistake is saying$$(v_{1,x},v_{1,y})= (v_0\cos{\theta},v_{0}\sin{\theta})$$This implies the x-component of ##m_1##’s momentum is unchanged by the explosion – as if the original mass divided into two but the two parts didn’t move apart. But, for example, a significant release of energy in the explosion could make ##v_{1,x}## any value, whereas ##v_0\cos{\theta}## would be unaffected.

For information$$3(\cos{\theta},\sin{\theta})= (v_{1,x},v_{1,y}) + 2(v_{2,x},v_{2,y}) =(v_{1,x},v_{1,y}) + 2(v_{2,x},v_{1,y})$$should be$$3v_0(\cos{\theta},\sin{\theta})= (v_{1,x},v_{1,y}) + 2(v_{2,x},v_{2,y}) = (v_{1,x},v_{1,y}) + 2(v_{2,x},v_{1,y})$$(But this is just a typo'.)
 
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