Conservation of momentum in an oblique launch and projectile explosion

In summary, the conversation discusses a problem involving calculating using moment conservation and finding the center of mass reach for a larger mass. The participants also discuss using the center of horizontal momentum frame and the importance of considering the horizontal components of the velocity.
  • #1
TheGreatDeadOne
22
0
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
..
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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  • #2
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.
 
  • #3
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
 
  • #4
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}}##.
 
  • #5
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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Related to Conservation of momentum in an oblique launch and projectile explosion

What is conservation of momentum?

Conservation of momentum is a fundamental principle in physics which states that the total momentum of a closed system remains constant. This means that in any interaction or event, the total momentum before and after the event must be equal.

How does conservation of momentum apply to an oblique launch?

In an oblique launch, the projectile is launched at an angle to the horizontal. Conservation of momentum applies in this situation because the total momentum of the system (the projectile and the launcher) must remain constant. This means that the horizontal and vertical components of momentum must be conserved separately.

What happens to the momentum of a projectile during an explosion?

During an explosion, the momentum of the projectile changes. Initially, the projectile has a certain amount of momentum in a specific direction. As the explosion occurs, the projectile's momentum changes in both magnitude and direction due to the forces acting on it. However, the total momentum of the system (the projectile and the explosive material) remains constant.

How does the angle of an oblique launch affect the conservation of momentum?

The angle of an oblique launch affects the conservation of momentum by changing the direction and magnitude of the projectile's momentum. As the angle increases, the vertical component of momentum increases while the horizontal component decreases. However, the total momentum of the system remains constant.

What other factors can affect conservation of momentum in an oblique launch and projectile explosion?

Other factors that can affect conservation of momentum include air resistance, friction, and external forces such as gravity. These factors can alter the trajectory and speed of the projectile, but the total momentum of the system will still remain constant.

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