Respuesta :
Answer:
[tex]v_3=8.622\ m.s^{-1}[/tex]
Explanation:
Given:
- mass of the gadget, [tex]m=21.85\ kg[/tex]
- mass of fragment 1, [tex]m_1=6.42\ kg[/tex]
- mass of fragment 2, [tex]m_2=8.26\ kg[/tex]
- velocity of fragment 1, [tex]v_1=6.8\ m.s^{-1}[/tex]
- velocity of fragment 2, [tex]v_2=3.54\ m.s^{-1}[/tex]
- angle between the velocity 1 & 2, [tex]\theta=64\ ^{\circ}[/tex]
Now the mass of the third part:
[tex]m_3=m-(m_1+m_2)[/tex]
[tex]m_3=21.85-(6.42+8.26)[/tex]
[tex]m_3=7.17\ kg[/tex]
The momentum of the gadget before and after collision is equal:
[tex]m_3.v_3=m_1.v_1\cos\ \frac{\theta}{2} +m_2.v_2\cos \frac{\theta}{2}[/tex]
[tex]7.17\times v_3=(6.42\times6.8+8.26\times3.54)\times \cos \frac{64}{2}[/tex]
[tex]v_3=8.622\ m.s^{-1}[/tex]
Refer schematic for more clarity.

The speed of third fragment of the given gadget is 8.62 m/s.
The given parameters:
- total mass of the gadget, Mt = 21.85 kg
- mass of first fragment, m₁ = 6.42 kg
- mass of the second fragment, m₂ = 8.26 kg
- speed of the first fragment, v₁ = 6.8 m/s
- speed of the second fragment, v₂ = 3.54 m/s
- angle between the first and second fragment, θ = 64⁰
The mass of the third fragment is calculated as follows;
[tex]m_3 = 21.85-(6.42 + 8.26)\\\\m_3 = 7.17 \ kg[/tex]
Apply the principle of conservation of linear momentum to determine the speed of the third fragment as follows;
[tex]m_3v_3 = m_1v_1 cos(\frac{\theta}{2} ) \ + \ m_2v_2 cos(\frac{\theta}{2} )\\\\7.17v_3 = 6.42\times 6.8 \times cos(\frac{64}{2} ) \ + 8.26 \times 6.8 \times cos(\frac{64}{2} )\\\\7.17 v_3 = 61.82 \\\\v_3 = \frac{61.82}{7.17} \\\\v_3 = 8.62 \ m/s[/tex]
Thus, the speed of third fragment of the given gadget is 8.62 m/s.
Learn more about conservation of linear momentum here: https://brainly.com/question/22698801
