Elastic collision of a ball on a wire

In summary: Just before collision, the block is in rest, vi1=0, and the velocity of the ball is obtained from the equation vi2=sqrt(gL) where the length of wire is L=1.2 m. Calculate vi2 and plug in the values for the initial velocities in both equations....it is clear that the velocities of the ball and the block will not be the same after collision. ehildFrom...Just before collision, the block is in rest, vi1=0, and the velocity of the ball is obtained from the equation vi2=sqrt(gL) where the length of wire is L=1.2 m. Calculate vi2 and plug in the values for the initial velocities in both
  • #1
jahrollins
25
0

Homework Statement


A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest. It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, 1.60 kg and 2.40 kg, and the length of the wire is 1.20 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.


Homework Equations


.5m1v2 = mgh
m1v1 + m2v2 = m1vf1 + m2vf2


The Attempt at a Solution


So I've got part a ~
.5m1v2 = mgh
v2 = 2gh

and I'm lost on part b:
m1v1 + m2v2 = m1vf1 + m2vf2
since the final velocities of the ball and box are unknown how do I solve?
 
Physics news on Phys.org
  • #2
After collision, total momentum and the KE is conserved.
Write down equations for these two conservation laws and solve for the final velocities.
 
  • #3
jahrollins said:

Homework Statement


A ball is attached to one end of a wire, the other end being fastened to the ceiling. The wire is held horizontal, and the ball is released from rest. It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, 1.60 kg and 2.40 kg, and the length of the wire is 1.20 m. Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.


Homework Equations


.5m1v2 = mgh
m1v1 + m2v2 = m1vf1 + m2vf2

The Attempt at a Solution


So I've got part a ~
.5m1v2 = mgh
v2 = 2gh

and I'm lost on part b:
m1v1 + m2v2 = m1vf1 + m2vf2
since the final velocities of the ball and box are unknown how do I solve?

Hi, for a) The ball starts from rest and has potential energy. The potential energy is converted to kinetic energy at the bottom of the string, therefore: mgh = 0.5mv^2, you shud get v = sqrt2gh, where h = the wire legnth = 1.2m.

For b) since the ball collides with the block, initially at rest, you have to use the law of conservation of momentum and adding the block mass and the ball mass and using the speed in the above question as the initial velocity of only the ball: mava = v(mball +mblock), where mava = the speed and mass of the ball respectively from the above question. Therefore, Solve for v.

Hope this helps:)
 
  • #4
E=mc^84 said:
Hi, for a) The ball starts from rest and has potential energy. The potential energy is converted to kinetic energy at the bottom of the string, therefore: mgh = 0.5mv^2, you shud get v = sqrt2gh, where h = the wire legnth = 1.2m.

For b) since the ball collides with the block, initially at rest, you have to use the law of conservation of momentum and adding the block mass and the ball mass and using the speed in the above question as the initial velocity of only the ball: mava = v(mball +mblock), where mava = the speed and mass of the ball respectively from the above question. Therefore, Solve for v.

Hope this helps:)
In mava = v(mball +mblock)
why would the final velocity be the same for both the ball and the block?
 
  • #5
The velocities will not be the same after collision.

The collision is elastic, both the momentum and the energy is conserved during the collision.

[tex]m_1v_{i1}+m_2v_{i2}=m_1v_{f1}+m_2v_{f2}[/tex]

[tex]\frac{1}{2}m_1v_{i1}^2+\frac{1}{2}m_2v_{i2}^2=\frac{1}{2}m_1v_{f1}^2+\frac{1}{2}m_2v_{f2}^2[/tex]

Just before collision, the block is in rest, vi1=0, and the velocity of the ball is obtained from the equation vi2=sqrt(gL) where the length of wire is L=1.2 m. Calculate vi2 and plug in the values for the initial velocities in both equations.

ehild
 
  • #6
jahrollins said:
In mava = v(mball +mblock)
why would the final velocity be the same for both the ball and the block?

Because they are in contact with each other during the collision for a reallt small amount of time. That velocity cud = the ball or the block since they are both moving at such initial speeds. But, since you are concerned with only the the velocity of the ball, you use only that.
 
  • #7
E=mc^84 said:
Because they are in contact with each other during the collision for a reallt small amount of time. That velocity cud = the ball or the block since they are both moving at such initial speeds. But, since you are concerned with only the the velocity of the ball, you use only that.

You are wrong. We do not care what happens during that very short time while the colliding bodies are in contact: it is beyond point mechanics. Only the result counts. If the collision is elastic, the kinetic energy is conserved, it is the same before and after collision.

The velocities will be the same if the collision is totally inelastic.

ehild
 
  • #8
ehild said:
You are wrong. We do not care what happens during that very short time while the colliding bodies are in contact: it is beyond point mechanics. Only the result counts. If the collision is elastic, the kinetic energy is conserved, it is the same before and after collision.

The velocities will be the same if the collision is totally inelastic.

ehild

You are talking about what you "care about", this is what happens in the problem. Work out the problem and show me that i am wrong :)
 
  • #9
E=mc^84 said:
You are talking about what you "care about", this is what happens in the problem. Work out the problem and show me that i am wrong :)

Well, do you know what is elastic collision?

ehild
 
  • #10
ehild said:
The velocities will not be the same after collision.

The collision is elastic, both the momentum and the energy is conserved during the collision.

[tex]m_1v_{i1}+m_2v_{i2}=m_1v_{f1}+m_2v_{f2}[/tex]

[tex]\frac{1}{2}m_1v_{i1}^2+\frac{1}{2}m_2v_{i2}^2=\frac{1}{2}m_1v_{f1}^2+\frac{1}{2}m_2v_{f2}^2[/tex]

Just before collision, the block is in rest, vi1=0, and the velocity of the ball is obtained from the equation vi2=sqrt(gL) where the length of wire is L=1.2 m. Calculate vi2 and plug in the values for the initial velocities in both equations.

ehild
From those two, I came out with vi2 = vf2?
 
  • #11
No. Plug in the numbers and solve the system of equations. ehild
 

1. What is an elastic collision?

An elastic collision is a type of collision in which both kinetic energy and momentum are conserved. This means that the total energy of the system before and after the collision remains the same.

2. How does a ball behave when it collides with a wire?

When a ball collides with a wire, it will experience an elastic collision if the wire is rigid and the ball is perfectly round and smooth. This means that the ball will bounce off the wire with the same speed and direction that it had before the collision.

3. What factors affect the elasticity of the collision between a ball and a wire?

The elasticity of the collision between a ball and a wire is affected by the materials of the ball and wire, the speed and angle of approach of the ball, and the rigidity of the wire.

4. How is the speed of the ball after the collision calculated?

The speed of the ball after the collision can be calculated using the formula v2 = v1 * (m1-m2)/(m1+m2), where v1 is the initial speed of the ball, m1 is the mass of the ball, and m2 is the mass of the wire. This formula assumes that the wire is much more massive than the ball.

5. Can a ball experience a perfectly elastic collision on a wire?

No, it is not possible for a ball to have a perfectly elastic collision on a wire. This is because some energy is always lost due to friction and deformation of the materials involved in the collision, making it impossible for the total energy of the system to remain constant.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
34
Views
595
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
997
  • Introductory Physics Homework Help
Replies
4
Views
967
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top