Transfer of momentum problem?

  • Thread starter greswd
  • Start date
  • Tags
    Momentum
In summary, the conversation discusses an interesting puzzle where a ball at rest collides with a moving ball in a one-dimensional space. The puzzle aims to prove that the ratio of the velocities of the two balls after the collision is less than 2. Using momentum and energy conservation equations, it can be shown that the ratio is indeed less than 2.
  • #1
greswd
764
20
Here's an interesting puzzle:

The whole scenario takes place in one dimension of space.

Ball B is at rest. Ball A moves with momentum [tex]p_a[/tex]

Ball A makes a perfectly elastic head-on collision with Ball B. Ball B moves off with momentum [tex]p_b[/tex]


Prove that [tex]|\frac{p_b}{p_a}|<2[/tex]
 
Physics news on Phys.org
  • #2
greswd said:
Here's an interesting puzzle:

The whole scenario takes place in one dimension of space.

Ball B is at rest. Ball A moves with momentum [tex]p_a[/tex]

Ball A makes a perfectly elastic head-on collision with Ball B. Ball B moves off with momentum [tex]p_b[/tex]Prove that [tex]|\frac{p_b}{p_a}|<2[/tex]

Momentum conservation :
[tex]{p_a} = {p_a}' + {p_b}[/tex]
[tex]1- \frac{{p_a}'}{p_a} = \frac{p_b}{p_a}[/tex]Energy conservation (Ball B had intially no KE, so speed of ball A cannot increase):
[tex]|\frac{{p_a}'}{p_a}| < 1[/tex]
 
Last edited:
  • #3
A.T. said:
[tex]|\frac{{p_a}'}{p_a}| < 1[/tex]
All true, but I don't see how that gets you to the answer required.
Using energy conservation more completely:
[tex]\frac{{p_a}'^2}{m_a}+\frac{{p_b}^2}{m_b}=\frac{{p_a}^2}{m_a}[/tex]
Combining with momentum eqn. we get
[tex]\frac{p_b}{p_a}=\frac{2m_b}{m_a+m_b}<2[/tex]
 
  • #4
haruspex said:
All true, but I don't see how that gets you to the answer required.
You have to combine the two of course:
[tex]1- \frac{{p_a}'}{p_a} = \frac{p_b}{p_a}[/tex]
[tex]|\frac{{p_a}'}{p_a}| < 1[/tex]
This leads to:
[tex]0 < \frac{p_b}{p_a} < 2[/tex]

haruspex said:
Using energy conservation more completely:
[tex]\frac{{p_a}'^2}{m_a}+\frac{{p_b}^2}{m_b}=\frac{{p_a}^2}{m_a}[/tex]
Combining with momentum eqn. we get
[tex]\frac{p_b}{p_a}=\frac{2m_b}{m_a+m_b}<2[/tex]

Yes, the clean & complete way is to derive the ratio as function of the masses. Mine was just showing that the ratio is < 2.
 
  • #5


This is a classic example of a transfer of momentum problem, where the momentum of one object is transferred to another object through a collision. In this case, we have two balls, A and B, with different initial momenta, p_a and 0 respectively.

After the collision, Ball B gains momentum p_b while Ball A loses all of its momentum. The key to solving this puzzle is to use the conservation of momentum principle, which states that the total momentum of a closed system remains constant.

In this case, the closed system is the two balls, and the total momentum before the collision is p_a + 0 = p_a. After the collision, the total momentum is p_b + 0 = p_b. Therefore, we can write the equation p_a = p_b.

To prove that |\frac{p_b}{p_a}| < 2, we can rewrite the equation as \frac{p_b}{p_a} < 2. This means that the ratio of the final momentum of Ball B to the initial momentum of Ball A must be less than 2.

To understand why this is true, let's consider the extreme case where Ball A has a very large initial momentum, close to infinity. In this scenario, the final momentum of Ball B will also be very large, but it will never be larger than twice the initial momentum of Ball A. This is because the conservation of momentum principle dictates that the total momentum of the system cannot increase.

In other words, the maximum possible value for the final momentum of Ball B is 2 times the initial momentum of Ball A, which proves the inequality \frac{p_b}{p_a} < 2.

In conclusion, the transfer of momentum problem can be solved using the conservation of momentum principle, and we can prove that the final momentum of Ball B will always be less than twice the initial momentum of Ball A. This is an interesting puzzle that demonstrates the fundamental principles of physics and the importance of understanding momentum in collisions.
 

1. What is the transfer of momentum problem?

The transfer of momentum problem is a physics concept that involves the exchange of momentum between two objects. It occurs when there is a collision or interaction between the objects, resulting in a change in their velocities.

2. How is momentum transferred between objects?

Momentum can be transferred between objects through either direct contact or through a force acting at a distance. In direct contact, the objects exchange momentum through the collision or interaction. In the case of a force acting at a distance, the objects can exchange momentum through the transfer of energy.

3. What factors affect the transfer of momentum?

The transfer of momentum between objects is affected by several factors, including the mass and velocity of the objects, the angle and direction of the collision or interaction, and the presence of external forces such as friction.

4. How is the transfer of momentum calculated?

The transfer of momentum can be calculated using the equation p = m * v, where p is the momentum, m is the mass of the object, and v is the velocity. It can also be calculated by comparing the initial and final momentum of the objects involved in the interaction.

5. Why is the transfer of momentum important?

The transfer of momentum is important in understanding and predicting the motion of objects. It is also a key concept in many fields of science, including physics, engineering, and astronomy. Additionally, it plays a crucial role in real-world applications such as car crashes, sports, and rocket propulsion.

Similar threads

  • Mechanics
2
Replies
53
Views
2K
Replies
30
Views
1K
Replies
9
Views
1K
Replies
3
Views
933
Replies
4
Views
1K
Replies
1
Views
746
Replies
5
Views
952
Replies
14
Views
7K
  • Mechanics
3
Replies
80
Views
10K
Replies
34
Views
1K
Back
Top