Conservation of momentum in an oblique launch and projectile explosion

[TheGreatDeadOne]

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:
$$t_1=t_2$$
$$\frac{R_1}{v_{1,x}}=\frac{R_2}{v_{2,x}}$$
$$R_2=\frac{R_1 v_{2,x}}{v_{1,x}}=\frac{3v_0^2}{g}$$

 

[PeroK]

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.

 

[TheGreatDeadOne]

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

 

[PeroK]

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 $\cos \theta = \frac 1 {\sqrt{2}}$.

 

[Steve4Physics]

$$ 3(\cos{\theta},\sin{\theta})= (v_{1,x},v_{1,y}) + 2(v_{2,x},v_{2,y})
=(v_{1,x},v_{1,y}) + 2(v_{2,x},v_{1,y})$$
$$ \therefore v_0 \sin{\theta}=v_{1,y}$$
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'.)