Classical mech. - inelastic collision

In summary, an inelastic collision is a type of collision in classical mechanics where kinetic energy is not conserved, although momentum is conserved. During such collisions, the colliding objects may deform or generate heat, leading to a loss of kinetic energy. The total momentum before and after the collision remains the same, but the total kinetic energy decreases. Inelastic collisions can be perfectly inelastic, where the objects stick together post-collision, or partially inelastic, where they separate but do not conserve kinetic energy fully.
  • #1
areverseay
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1
Homework Statement
A particle A of mass 2m travelling with speed u undergoes an inelastic collision with a
stationary particle B of mass 10m. After the collision, A travels with a speed vA at 90 degrees to its original direction of motion and B travels with speed vB at an angle θ = arcsin(3/5) to the original direction of motion of A. Express vA and vB in terms of u; what fraction of kinetic energy is lost as a result of the collision?
Relevant Equations
p = mv
KE = (p^2)/2m
vA = 3u/4 and vB = u/4, and 1/8 KE is lost. I can't get to these answers however: for the first part, I got to u = vA + 3vB using conservation of momentum, and the fact that particle B is at an angle, hence I would think its momentum should be 10mvBsin(arcsin(3/5)). Doing the same for A with angle sin90 gives 2mu = 2mvA + 6mvB, hence the equation I first listed, but I don't know how to isolate vA and vB in terms of u. For the second part, even with vA and vB in terms of u, I can't get to 1/8 lost: using KE = (p^2)/2m and summing the KE for particles A and B, I get (54mu^2)/80.
 
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  • #2
areverseay said:
2mu = 2mvA + 6mvB
How do you figure that? u and vA are at right angles.
 
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  • #3
areverseay said:
sorry for 3 questions
Per forum guidelines, please separate into three threads,
 
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  • #4
haruspex said:
Per forum guidelines, please separate into three threads,
Thanks for this, I've made this post just the first question.

In retrospect it doesn't really make much sense, since as you say u and vA are perpendicular, but I'm not sure how else to get anything for vA - if I were to use cos90, the particle would be stationary, which isn't true, and I don't know how else to get an expression.
 
  • #5
areverseay said:
Thanks for this, I've made this post just the first question.

In retrospect it doesn't really make much sense, since as you say u and vA are perpendicular, but I'm not sure how else to get anything for vA - if I were to use cos90, the particle would be stationary, which isn't true, and I don't know how else to get an expression.
No, it would not be stationary, it just wouldn’t have a velocity component in its original direction. That is perfectly possible in general.
The component of vB in that direction is the cos of the angle, not the sine.
 
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  • #6
areverseay said:
Thanks for this, I've made this post just the first question.

In retrospect it doesn't really make much sense, since as you say u and vA are perpendicular, but I'm not sure how else to get anything for vA - if I were to use cos90, the particle would be stationary, which isn't true, and I don't know how else to get an expression.
You must write conservation of momentum for each velocity component before and immediately after collision. It applies to both horizontal and vertical directions. The mass ##2m## leaves the impact with only a vertical momentum component (w.r.t. to the coordinates I chose), which balances the vertical momentum component of the ##10m## mass leaving the impact. Apply this concept to the horizontal direction too. You can get the velocities of each mass in terms of ##u##, which can then be used to determine the energy loss.
 
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  • #7
erobz said:
You must write conservation of momentum for each velocity component before and immediately after collision. It applies to both horizontal and vertical directions. The mass ##2m## leaves the impact with only a vertical momentum component (w.r.t. to the coordinates I chose), which balances the vertical momentum component of the ##10m## mass leaving the impact. Apply this concept to the horizontal direction too. You can get the velocities of each mass in terms of ##u##, which can then be used to determine the energy loss.
Both of these replies are very helpful, but I'm still stuck on the horizontal component. So, the vertical component of the momentum of B is the same as the original momentum of A, since there is no vertical component in the new momentum of A, as cos90 = 0 / they are perpendicular. But how do you get the horizontal component? Since there is none in the original momentum, I guess the horizontal component of both new A and B sums to 0, but how do I link that back to u?
 
  • #8
areverseay said:
the horizontal component of both new A and B sums to 0
Yes
areverseay said:
how do I link that back to u?
Write the equations and that may become apparent.
 
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  • #9
areverseay said:
Both of these replies are very helpful, but I'm still stuck on the horizontal component. So, the vertical component of the momentum of B is the same as the original momentum of A,
No.

Please write conservation of momentum for the vertical direction, ##p_{before} ~\hat{\text{j}} = p_{after} ~\hat{\text{j}} ##

What do you get for this expression? Is there any momentum in the vertical direction before the collision? What are the vertical components of each masses momentum after the collision?
 
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  • #10
erobz said:
No.
Are you making an assumption about which is vertical and which is horizontal in the OP's view?
 
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I don't know. Where is the diagram that labels these vectors? We are talking about vertical momentum. I take that to mean up and down on the screen, and left and right as horizontal. I said before that is with respect to the coordinates I chose. Of course the OP's problem could be oriented in anyway.

To the OP: orient it however you want, just give an indication as to the directions you chose? I diagram would be helpful in either case here. I'm not trying to confuse you, but we better first agree on what is up/down, left/right ...
 
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  • #12
erobz said:
I don't know. Where is the diagram that labels these vectors? We are talking about vertical momentum. I take that to mean up and down on the screen, and left and right as horizontal. I said before that is with respect to the coordinates I chose. Of course the OP's problem could be oriented in anyway.

To the OP: orient it however you want, just give an indication as to the directions you chose? I diagram would be helpful in either case here.
Yes, I think we had different views of which orientation everything was. But, I've figured out the correct answers for all of the parts, so it's fine. Thanks for both of your help, explicitly writing the equations for conservation of momentum did indeed help a lot!
 
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  • #13
areverseay said:
Yes, I think we had different views of which orientation everything was. But, I've figured out the correct answers for all of the parts, so it's fine. Thanks for both of your help, explicitly writing the equations for conservation of momentum did indeed help a lot!
That's good news...time for bed!
 
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FAQ: Classical mech. - inelastic collision

What is an inelastic collision in classical mechanics?

An inelastic collision is a type of collision in classical mechanics where the kinetic energy of the system is not conserved. Instead, some of the kinetic energy is transformed into other forms of energy, such as heat, sound, or deformation energy. However, the total momentum of the system is conserved.

How is momentum conserved in an inelastic collision?

In an inelastic collision, the total momentum of the system before the collision is equal to the total momentum of the system after the collision. This is due to the principle of conservation of momentum, which states that in the absence of external forces, the total momentum of a closed system remains constant. Mathematically, this can be expressed as \(m_1 \cdot v_1 + m_2 \cdot v_2 = m_1 \cdot v_1' + m_2 \cdot v_2'\), where \(m\) and \(v\) represent the masses and velocities of the colliding objects before and after the collision, respectively.

What is the difference between elastic and inelastic collisions?

The main difference between elastic and inelastic collisions lies in the conservation of kinetic energy. In an elastic collision, both momentum and kinetic energy are conserved, meaning no kinetic energy is lost to other forms of energy. In contrast, in an inelastic collision, momentum is conserved, but kinetic energy is not; some of the kinetic energy is converted into other forms of energy.

Can you provide an example of an inelastic collision?

A common example of an inelastic collision is a car crash. When two cars collide, they often crumple and generate heat and sound, indicating that some of the kinetic energy has been converted into other forms of energy. Despite the loss of kinetic energy, the total momentum of the cars before and after the collision remains conserved.

How do you calculate the final velocities in a perfectly inelastic collision?

In a perfectly inelastic collision, the colliding objects stick together after the collision, moving with a common velocity. To find this final velocity, you can use the conservation of momentum. If \(m_1\) and \(m_2\) are the masses of the objects, and \(v_1\) and \(v_2\) are their initial velocities, the final velocity \(v_f\) can be calculated using the equation: \(v_f = \frac{m_1 \cdot v_1 + m_2 \cdot v_2}{m_1 + m_2}\). This equation ensures that the total momentum before and after the collision is the same.

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