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# Impact

An analysis of the effects of an impact between two bodies
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## Overview

If a body of weight $\inline&space;\displaystyle&space;W_1$ overtakes a body of weight $\inline&space;\displaystyle&space;W_2$ and their velocities are $\inline&space;\displaystyle&space;u_1\;\;\;and\;\;\;u_2$ before impact and $\inline&space;\displaystyle&space;v_1\;\;\;and\;\;\;v_2$ Then if the reaction between the bodies is R and the impulse lasts a short time r:-

$\frac{W_2\,v_2}{g}&space;-&space;\frac{W_2\,u_2}{g}&space;=&space;\int_{0}^{r}\;R\;dt$
and
$\frac{W_1\,v_1}{g}\;-&space;\frac{W_1\,u_1}{g}\;=&space;-&space;\int_{0}^{r}R\,dt$

Because the force on $\inline&space;\displaystyle&space;W_1$ is in the opposite direction of motion, its impulse in the direction of motion is negative.

Adding the above two equations gives:
$\frac{W_1\,v_1}{g}&space;+&space;\frac{W_2\,v_2}{g}&space;=&space;\frac{W_1u_1}{g}&space;+&space;\frac{W_2\,u_2}{g}$

This equation states that the total momentum after impact is the same as the momentum before. This is the principle of The Conservation of Momentum.

### Example 1

One end of a chain of length 2l is fixed. The other end is held level with it and then let go. If the weight of the whole chain is W, find the tension at the lowest point when the free end has dropped a distance x. Also find the the tension at the fixed end when the whole chain has just become straight. Explain how it is possible for this to be greater than W.

In a time $\inline&space;\displaystyle&space;\delta&space;t$ the free end drops a distance $\inline&space;\displaystyle&space;\delta&space;x$ and so a length $\inline&space;\displaystyle&space;\frac{1}{2}\delta&space;x$ near the lowest point of the chain, which was moving at velocity $\inline&space;\displaystyle&space;\dot{x}$ at the beginning of $\inline&space;\displaystyle&space;\delta&space;t$, comes to rest at the end of $\inline&space;\displaystyle&space;\delta&space;t$.

Since the weight of a length $\inline&space;\displaystyle&space;\frac{1}{2}\delta&space;x$ is $\inline&space;\displaystyle&space;\frac{W}{2l}\times\frac{1}{2}\delta&space;x$ . It loses momentum $\inline&space;\displaystyle&space;\frac{W}{2gl}\times\frac{1}{2}\delta&space;x&space;\times\dot{x}$ ( to the first order) in a time $\inline&space;\displaystyle&space;\delta&space;t$

Therefore momentum is destroyed at the following rate:-

$\lim_{\delta&space;t\rightarrow&space;0}\left(\frac{W}{2gl}\times\frac{1}{2}\delta&space;x\times\dot{x}&space;\right)\div&space;\delta&space;t&space;=&space;\frac{W}{4gl}\times\dot{x}^2$

But $\inline&space;\displaystyle&space;\dot{x}$ is the velocity due to falling a distance x under gravity i.e.$\inline&space;\displaystyle&space;\dot{x}^2&space;=&space;2gx$. Hence the momentum of the element is destroyed at a rate of $\inline&space;\displaystyle&space;\frac{Wx}{2l}$. This must be equal to the force on the element in the opposite direction to its motion,. This is the t4ension in the chain at the lowest point. Hence the tension of the chain at the lowest point is $\inline&space;\displaystyle&space;\frac{Wx}{2l}$

The Tension at the fixed point is obtained by adding the weight of the stationary part of the chain.
$\frac{W}{2l}\times(l&space;+&space;\frac{1}{2}x)$
.

Hence the total weight is given by:
$\frac{W(2l&space;+&space;3x)}{4l}$

Immediately before the chain straightens the tension at the top is 2W and that at the bottom W. Immediately it has straightened the tension at the bottom must be zero. It seems paradoxical that the tension at the top can suddenly drop from 2W to W, which is the force just necessary to keep the chain in static equilibrium. The explanation is that the peg from which the chain hangs becomes more and more compressed as the chain falls until it exerts a force of 2W on the chain. Then the compression decreases and the chain oscillates up and down until the movement is damped out and the force exerted by the peg settles at W. In general the peg has very little elasticity and the oscillation is damped out so quickly as to be unnoticeable.

### Example 2

A Pile Driver has a hammer of weight W an it is used to drive a pile of weight w. The hammer is dropped a distance H and the pile is driven a distance h. Find the average resistance of the ground, assuming that the pile and hammer have the same velocity after the impact. Calculate the loss of energy at the impact.

The velocity of the hammer just before impact is $\inline&space;\displaystyle&space;\sqrt{2\,g\,H}$. If the velocity of the hammer and pile after impact is v then by the principle of the conservation of Momentum.

$\frac{W&space;+&space;w}{g}\;v&space;=&space;\frac{W\sqrt{2gH}}{g}$

Also if the resistance of the ground is R, then the downward force on the pile and hammer is $\inline&space;\displaystyle&space;(W&space;+&space;w)&space;-&space;R$ and hence the equation of energy for their motion after the impact is:-

$-&space;\frac{(W&space;+&space;w)}{2g}\;v^2&space;=&space;\left\{(W&space;+&space;w)&space;-&space;R&space;\right\}h$

$\therefore\;\;\;\;R&space;=&space;(W&space;+&space;w)&space;+&space;\frac{(W&space;+&space;w)\;v^2}{2gh}$

Squaring both sides of equation (5)

$(W&space;+&space;w)^2\times&space;v^2&space;=&space;W^2\times&space;2gH$

$Hence\;\;\;\;\frac{(W&space;+&space;w)\times&space;v^2}{2g}&space;=&space;\frac{W^2\times&space;H}{(W&space;+&space;w)}$

Substituting equation(9) in equation (7) and eliminating v

$R&space;=&space;(W&space;+&space;w)&space;+&space;\frac{W^2\,H}{(W\;\;+\:w)h}$

The Energy lost at the Impact is the difference between the initial Potential Energy of the hammer (W H) and the kinetic Energy of the Hammer and Pile immediately after impact.

$Energy\;lost&space;=&space;W\,H&space;-&space;\frac{(W&space;+&space;w)\;v^2}{2g}$

Again using equation (9) this can be re-written to eliminate v

$Energy\;lost&space;=&space;W\,H&space;-&space;\frac{W^2\,H}{(W&space;+&space;w)}&space;=&space;\frac{W\,w\,H}{(W&space;+&space;w)}$