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A(n) 18 kg object, initially at rest in free space, "explodes" into three segments. The masses of two of these segments are both 7 kg and their velocities are 4.3 m/s. The angle between the direction of motion of these segments is 74 ◦ . What is the speed of the third segment?

Respuesta :

Answer:

[tex]v_3 = 12 m/s[/tex]

Explanation:

As we know that during explosion of the object there is no external force on the system

So total momentum of the system will always remain conserved

so we will have

[tex]m_1 v_1 + m_2v_2 + m_3 v_3 = 0[/tex]

so here we can say

[tex]\vec v_3 = - \frac{(m_2 \vec v_2 + m_1 \vec v_1)}{m_3}[/tex]

so here we know that

[tex]m_1v_1 = m_2 v_2 = (7 )(4.3)[/tex]

[tex]m_1v_1 = 30.1 kg m/s[/tex]

now we know that

[tex]\vec v_3 = - \frac{2mv cos\frac{\theta}{2}}{m_3}[/tex]

[tex]\vec v_3 = - \frac{2(30.1)cos\frac{74}{2}}{4}[/tex]

[tex]v_3 = 12 m/s[/tex]

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