- #1
TheGreatDeadOne
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- Homework Statement
- A projectile of mass ##m## is fired with an initial velocity of modulus ##v_0## at an elevation angle of ##45^\circ##. The projectile explodes in the air into two pieces of masses ##m/3## and ##2m/3##. The pieces continue moving in the same plane as the entire projectile and reach the ground together. The smaller piece falls at a distance of ##R_{1}=(3v_{0}^{2})/(2g)## from the launch point. Determine the range of the larger piece. Disregard air resistance
- Relevant Equations
- Listed below
I've already solved this problem using another resource (just get the coordinate of the range of the center of mass and from there, get it for the larger mass ##R_{2}=(3v_{0}^{2})/(4g))##:
Range CM: $$R_{(CM)} = \frac{v_{0}^2 sin{2\theta}}{2g}=\frac{v_{0}^{2}}{2}$$
then:
$$ R_{(CM)}= \frac{v_{0}{2}}{2} = \frac{m_{1}R_{1}+m_{2}R_{2}}{M} \rightarrow \frac{v_{0}{2}}{2} = \frac{1v_{0}}{2g} + \frac{2R_{2}}{3} $$
$$\therefore R_{2}=\frac{3v_{0}^2}{4g} $$
Trying to solve using the conservation of linear momentum, we have:
$$m\vec v_{0} = \frac{1}{3}m\vec v_{1} + \frac{2}{3}m\vec v_{2}$$
In the most general possible case, we know that the fragments can also perform an oblique launch. As we know the angle associated with these new oblique launches, we also determine the horizontal component associated with each fragment. So:
$$ v_{1x} =
\frac{1}{3}v_{0}\cos(\theta)
\quad \text{and} \quad v_{2x} =
\frac{2}{3}v_{0}\cos(\theta)
$$
In addition, we can equalize the flying time intervals of each fragment by the horizontal component of each fragment (uniform movement), but this is where I get stuck:
$$ t_{1} = t_{2} \longrightarrow \frac{R_{1}}{v_{1x}}=\frac{R_{2}}{v_{2x}} \longrightarrow R_{2}= \frac{R_{1}v_{2x}}{v_{1x}}=\frac{3v_{0}^2}{g} (????????)$$
What could I be doing wrong here? I'm sure I'm missing something very obvious.
Range CM: $$R_{(CM)} = \frac{v_{0}^2 sin{2\theta}}{2g}=\frac{v_{0}^{2}}{2}$$
then:
$$ R_{(CM)}= \frac{v_{0}{2}}{2} = \frac{m_{1}R_{1}+m_{2}R_{2}}{M} \rightarrow \frac{v_{0}{2}}{2} = \frac{1v_{0}}{2g} + \frac{2R_{2}}{3} $$
$$\therefore R_{2}=\frac{3v_{0}^2}{4g} $$
Trying to solve using the conservation of linear momentum, we have:
$$m\vec v_{0} = \frac{1}{3}m\vec v_{1} + \frac{2}{3}m\vec v_{2}$$
In the most general possible case, we know that the fragments can also perform an oblique launch. As we know the angle associated with these new oblique launches, we also determine the horizontal component associated with each fragment. So:
$$ v_{1x} =
\frac{1}{3}v_{0}\cos(\theta)
\quad \text{and} \quad v_{2x} =
\frac{2}{3}v_{0}\cos(\theta)
$$
In addition, we can equalize the flying time intervals of each fragment by the horizontal component of each fragment (uniform movement), but this is where I get stuck:
$$ t_{1} = t_{2} \longrightarrow \frac{R_{1}}{v_{1x}}=\frac{R_{2}}{v_{2x}} \longrightarrow R_{2}= \frac{R_{1}v_{2x}}{v_{1x}}=\frac{3v_{0}^2}{g} (????????)$$
What could I be doing wrong here? I'm sure I'm missing something very obvious.
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