• #1
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Main Question or Discussion Point

Introduction
Collisions are very much a stock item in any school physics curriculum and students are generally taught about the use of the principles of conservation of momentum and energy for solving simple collision problems in one dimension. In this article we will be examining a very common type of collision problem: the inelastic or “collide and coalesce” collision. (See Note 1 below: a review comment on the definition of an inelastic collision)  However we will be following a somewhat unconventional approach which – notwithstanding – is very much in step with a key underlying physical law namely Newton’s third law upon which conservation of momentum rests. We develop some easy to use formulae for the solution of inelastic collisions and show that these may be easily modified to solve both perfectly elastic and ‘semi-elastic’ collision problems.
Conservation of Momentum re-arranged
The basic equation for conservation of momentum in a straightforward...
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Answers and Replies

  • #2
63
5
However a much simpler method arises from the recognition that the physics of the above perfectly inelastic collision is mathematically equivalent to a collision between a single body having a mass corresponding to the reduced mass (and velocity equal to the relative velocity Δv) and a wall of infinite mass.

Source https://www.physicsforums.com/insights/an-alternative-approach-to-solving-collision-problems/
Did you mean the reduced mass is equivalent to ##\Delta m## or ##m_2 - m_1## where ##m_2## is mass of body 2 and ##m_1## is mass of body 1?
 
  • #3
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Did you mean the reduced mass is equivalent to ##\Delta m## or ##m_2 - m_1## where ##m_2## is mass of body 2 and ##m_1## is mass of body 1?
No - the formula for "reduced mass" or perhaps "harmonic mass" is given by the formula: ##\frac{m_1m_2}{m_1+m_2}## or ##{\left(\frac{1}{m_1}+\frac{1}{m_2}\right)}^{-1}##
 
  • #4
pbuk
Science Advisor
Gold Member
1,282
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I think you need a better example to show the advantage of your method: you take 5 steps to get to the solution to 4.4 but the 'classic' solution takes only 2 using only ## \mu = \Sigma mv ##:
  1. Initial momentum (taking East +ve) is ## 1,000 \times 40 - 5,000 \times 20 = 40,000 - 100,000 = -60,000 \mathrm{\ kg\ m\ s^{-1}} ##
  2. Conservation of momentum => final velocity given by ##(1,000 + 5,000) v = -60,000 \implies v = -10 \mathrm{\ m\ s^{-1}} ##

And for 4.5 it's 3 vs 1 using only ## F = ma ## and ## a = \frac{\Delta v}{\Delta t} ##:
  1. The force is given by ## F = ma = m \dfrac{\Delta v}{\Delta t} = 1,000 \dfrac{40 - (-10)}{0.5} = 1,000 \times \dfrac{50}{0.5} = 100,000 \mathrm{\ N} ##

Or am I missing something?
 
  • #5
478
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I think you need a better example to show the advantage of your method: you take 5 steps to get to the solution to 4.4 but the 'classic' solution takes only 2 using only ## \mu = \Sigma mv ##:
  1. Initial momentum (taking East +ve) is ## 1,000 \times 40 - 5,000 \times 20 = 40,000 - 100,000 = -60,000 \mathrm{\ kg\ m\ s^{-1}} ##
  2. Conservation of momentum => final velocity given by ##(1,000 + 5,000) v = -60,000 \implies v = -10 \mathrm{\ m\ s^{-1}} ##

And for 4.5 it's 3 vs 1 using only ## F = ma ## and ## a = \frac{\Delta v}{\Delta t} ##:
  1. The force is given by ## F = ma = m \dfrac{\Delta v}{\Delta t} = 1,000 \dfrac{40 - (-10)}{0.5} = 1,000 \times \dfrac{50}{0.5} = 100,000 \mathrm{\ N} ##

Or am I missing something?
Thanks for taking a look - the first example was intended to be simple and detailed so as to clearly indicate the steps in the method. Once you are used to it you will solve 4.4 as follows:

$$v_f=\frac{P_1-\Delta P}{m_1} $$
$$ \Delta P=\mu \Delta v= \frac{1000\times5000}{1000+500}\times(40 - (-20))=50000 Ns$$
$$v_f=\frac{40\times1000-50000}{1000}=-10ms^{-1}$$

And for 4.5 $$F_{truck\; on\; car}=\frac{-\Delta P}{\Delta t}=\frac{-50000}{0.5}=-100000N$$

You might like to try the following from an MIT problem set (Physics 8.01 Fall 2012).

MIT Problem 5.jpg

20th Feb 2020: Added above problem as another worked example in the article.
 
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  • #6
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The use of the reduced mass is restricted to the scenario of frontal approach, that is the reason why this formula is not widely used to explain the phenomenon, since the use of general expressions is in general preferable.
 
  • #7
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Not restricted by anything different from conservation of momentum and conservation of energy. Will prepare an example shortly but thanks for the comment all the same. Prompts a bit of work on my part!

You might like to try Problem 6 from the same MIT Collisions Problem set as Problem 5 above. (PS: if anyone from MIT views this, please advise on how I should reference this particular problem set which I downloaded some time ago but can no longer find online. It's Problem Set 7 dated Fall 2012).

Problem 6.jpg

As before I'll eventually get round to adding this as another worked example using the alternate method.
 
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