Inelastic collision and circular motion

In summary, the conversation discusses the problem of finding the minimum initial speed of a 5.00kg dart that strikes a 20.00kg lead sphere hanging from a thin wire, in order for the combination to make a complete circular loop. The equations used include those for uniform circular motion and conservation of energy. After some calculations, the minimum initial speed of the dart was determined to be 65.5m/s.
  • #1
seanpk92
6
0

Homework Statement


A 20.00kg lead sphere is hanging from a hook by a thin wire 3.50m long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a 5.00kg dart that embeds itself in the lead sphere. What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision?


Homework Equations


w = omega
a = alpha
[STRIKE]o[/STRIKE] = theta
a = (rw)^2
w = w(i) + at
[STRIKE]o[/STRIKE] = [STRIKE]o[/STRIKE](i) + w(i)t + .5at^2
m(a1)v(a1) + m(b1)v(b1) = (m[a] + m)*v(2)
acc(rad) = v^2/R

The Attempt at a Solution


m(a1) = 20kg m(b1) = 5kg
v(a1) = 0m/s v(b1) = ?m/s r = 3.5m

20*0 + 5*? = 25*v(2)


I have no idea what to do with this. Like so I find velocity first?
 
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  • #2
In order to make a circular loop, what minimum speed must the sphere+dart have at the top of the loop?
 
  • #3
Should I use the uniform circular motion equation for the Vmin at the top of the loop
 
  • #4
seanpk92 said:
Should I use the uniform circular motion equation for the Vmin at the top of the loop


I figured it out, I think. I used:
mg = m(v^2/R)
v = sqrt(gR)
K(1) + U(g1) = K(2) + U(g2)

U(g1) = 0
y(2) = 2R
drop the masses since they cancel out
.5v(c)^2 = .5v(2)^2 + 2gR
then plug v(c) back into COM and solve.
I got 65.5m/s
 
  • #5
You're getting there. What value did you get for the velocity of the sphere plus dart at the top of the loop? How about at the bottom?

Your value for the velocity of the dart looks a tad high to me. Perhaps you could show your calculation.
 

1. What is an inelastic collision?

An inelastic collision is a type of collision between two objects where there is a loss of kinetic energy. This means that the total kinetic energy of the system before and after the collision is not conserved. In an inelastic collision, the objects stick together and move with a common velocity after the collision.

2. How is momentum conserved in an inelastic collision?

In an inelastic collision, momentum is still conserved even though kinetic energy is lost. This means that the total momentum of the system before the collision is equal to the total momentum after the collision. The objects may have different velocities before and after the collision, but the total momentum remains the same.

3. What is circular motion?

Circular motion is the movement of an object along a circular path. The object travels at a constant speed but changes direction continuously, resulting in a circular motion. This can be seen in objects moving in a circular orbit, like planets around the sun, or in objects moving in a circular track, like cars on a racetrack.

4. How is circular motion related to inelastic collision?

Inelastic collisions can occur in circular motion. For example, a car going around a circular track may collide with a barrier, causing an inelastic collision. The car will then continue to move in a circular motion but with a different velocity due to the loss of kinetic energy in the collision.

5. What is the difference between elastic and inelastic collisions?

In an elastic collision, both kinetic energy and momentum are conserved. This means that the total kinetic energy of the system before and after the collision is the same, and the total momentum is also conserved. In an inelastic collision, only momentum is conserved, and there is a loss of kinetic energy. Additionally, in an elastic collision, the objects bounce off each other, while in an inelastic collision, the objects stick together and move with a common velocity.

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