Respuesta :
Explanation:
It is known that motion of objects on a vertical is an example of non-uniform motion.
When an object is at the highest point of the circle then for crossing highest point the centripetal force balances the weight of the object.
Therefore, [tex]\frac{mv^{2}}{r} = mg[/tex]
At the highest point of the circle, the minimum speed is as follows.
v = [tex]\sqrt{rg}[/tex]
= [tex]\sqrt{2.80 \times 9.8}[/tex]
= 5.23 m/s
As the sphere falls from highest to lowest point of circle as follows. According to the law of conservation of energy,
[tex]K.E_{lowest} = K.E_{highest} + P.E_{highest}[/tex]
Expression for potential energy is as follows.
[tex]P.E_{highest} = mgh[/tex]
where, h = diameter of the circle (2r)
So, [tex]P.E_{highest} = mgh[/tex]
[tex]P.E_{highest} = mg(2r)[/tex]
[tex]\frac{1}{2}mu^{2} = \frac{1}{2}mv^{2} + mg(2r)[/tex]
where, u = velocity at the lowest time
Hence, the above equation will be as follows.
[tex]u^{2} = v^{2} + 4gr[/tex]
u = [tex]\sqrt{v^{2} + 4gr}[/tex]
= [tex]\sqrt{(5.23)^{2} + (4 \times 9.8 \times 2.80)}[/tex]
= 11.71 m/s
The dart is embedded into bullet and the collision is inelastic. From the law of conservation of momentum,
[tex]m_{2}v = (m_{1} + m_{2})u[/tex]
v = [tex]\frac{(m_{1} + m_{2})u}{m_{2}}[/tex]
= [tex]\frac{(20 + 5) \times 11.71}{5}[/tex]
= [tex]\frac{292.75}{5}[/tex]
= 58.55 m/s
Thus, we can conclude that the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision is 58.55 m/s.
After collision, the maximum speed of the dart will be "58.55 m/s".
Collision
According to the question,
Mass of lead sphere = 20.00 kg
Length of wire = 1.80 m
Mass of steel dart = 5.00 kg
We know the relation,
Maximum speed, v = [tex]\sqrt{rg}[/tex]
By substituting the values,
= [tex]\sqrt{2.80\times 9.8}[/tex]
= 5.23 m/s
By using law of conservation of energy,
→ [tex]K.E_{lowest} = K.E_{highest} + P.E_{highest}[/tex]
or,
→ [tex]P.E_{higher} = mgh[/tex]
= mg(2r)
[tex]\frac{1}{2}[/tex]mu² = [tex]\frac{1}{2}[/tex]mv² + mg(2r)
Now,
→ u² = v² + 4gr
u = [tex]\sqrt{v^2+4gr}[/tex]
= [tex]\sqrt{(5.23)^2 +(4\times 9.8\times 2.80)}[/tex]
= 11.71 m/s
By using conservation of momentum,
→ m₂v = (m₁ + m₂)u
or,
v = [tex]\frac{(m_1+m_2)v}{m_2}[/tex]
By putting the values,
= [tex]\frac{(20+5)\times 11.71}{5}[/tex]
= [tex]\frac{292.75}{5}[/tex]
= 58.55 m/s
Thus the response above is correct.
Find out more information about collision here:
https://brainly.com/question/7694106
