We will consider problems of impact and collision.

Body one is moving with speed $v_1$, and body two is moving with a speed of $v_2$.

They touch each other: Impact

Two bodies, one and two of mass $m_1$ and $m_2$ impact each other and move after the impact.

Impact - COLLISION

The bodies are approaching each other, can we find the final velocities of these bodies using the laws of mechanics.

Impact takes place over a very small period of time.

Impact force $\rightarrow$ large.

Impact $\rightarrow$ instantaneous impulsive force.

$i \rightarrow$ just before impact or pre impact.

$f \rightarrow$ position just after impact.

Because the time of impact is very small, we will assume position of bodies do not change.

$t \rightarrow$ tangent plane of the bodies.

In two dimension, $t$ will be a line.

Perpendicular to $t$ in the plane.

$n$ direction is the line of impact.

When we have two impacting bodies, we identify

t $\rightarrow$ as the plane of contact.

n $\rightarrow$ line of impact.

Smooth impact $\Rightarrow$ impact force on each body separately is along the line of impact.

Collision force only along $n$ for smooth impact.

There is no force in the tangent plane.

$v_{1 i}$: Velocity of body 1 before impact

$v_{2 i}$: Velocity of body 2 before impact

If $\overrightarrow{v_{1 i}}$ and $\vec{v}_{2 i}$ are only along $n$ direction, then the collision is head-on / direct collision.

How to decide if a collision is head-on or a direct collision?

Draw the tangent plane, $t \rightarrow (n \perp t)$.

Normally in a problem $\overrightarrow{v_{1 i}}, \overrightarrow{v_{2 i}}$, $m_1$ and $m_2$ are given.

We wish to find $\overrightarrow{v_{1 f}}$, and $\overrightarrow{v_{2 f}}$.

Impulse on body 1 = $\int F dt$.

Impulse on body 2 = -$\int F dt$.

If we apply Impulse momentum principle on body 1:

$\overrightarrow{v_{1 i}}$ is only along $n$.

$m_1 \overrightarrow{v_{1 i}}+\overrightarrow{I} {=} m_1 \overrightarrow{v_{1 f}}$

Initially momentum: $m_1 \overrightarrow{v_{1 i}}$

Finally momentum: $m_1 \overrightarrow{v_{1 f}}$

If $\overrightarrow{v_{1 i}}$ only along $n$ i.e., head on collision,

$t$ component of $\vec{v}_{l i}=0$

Impulse is also along $\hat{n}$,$\overrightarrow{v_{1 f}}$ is also along $\hat{n}$

$t$ component of $\overrightarrow{v}_{1 f}=0$.

Both the bodies will move with the same velocity just after collision.

$\vec{v}_{1 i} \rightarrow$ n component and t component $\neq 0$.

Impulsive force is only along n.

$t$ component of momentum of body 1 will not change, because impulse on the body 1 is only along $n$

$(v_ {1t}) _ i = (v_ {1t})_f$

Same arguments hold for body 2.

$\vec{v{ }_{2 i}}$ (intial velocity)

$\vec{v}{ }_{2 f}$ (final velocity)

Since Impulse is only along $\hat{n}$ :

$(v_ {2t}) _ i = (v_ {2t})_f$

Our unknowns are $v_{1 f}$ and $v_{2 f}$(which are along n).

$-m_2 v_2 i+{I} =m_2 v_{2 f}$

Alternate: Momentum equation in the $n$ direction for both the bodies together.

Initial momentum along $n$ direction for both the bodies: $m_1 v_{1 i}-m_2 v_{2 i}$

$v_{1 f}$ and $v_{2 f}$(positive along n).

Impulse for both the bodies as a system = 0

Momentum of both bodies along $x$ direction must be conserved.

$m_1 v_{1 i}-m_2 v_{2 i}=m_1 v_{1 f}+m_2 v_{2 f}$

This is one equation and there are two unknowns $v_{1 f}$ and $v_{2 f}$.

Need some extra information.

Coefficient of restitution 'e'

For elastic collision $e=1$.

Fully non-elastic or a plastic collision $e=0$.

Elastic collision: $e=1$

Kinetic energy of the two bodies together as system is conserved before and after collision.

If a collision is not elastic then (kinetic energy post collision) $<$ (kinetic energy pre collision).

Lost energy in non elastic collisions may be converted to sound / internal energy of the two bodies show up as heat.

Completely inelastic collision $e=0$

$v_{\text {1 f}}$ and $v_{2 f}$ in n direction are equal.

n component of $v_{1 f}$ = n component of $v_{2_f}$

e = Coefficient of Restitution

$e=\frac{- \text { (elative velocity of separation) }}{\text { (relative velocity of approach) }}$

Separation is always post impact and approach is pre-impact.

For given problem

Relative velocity of separation$=v_{\text {i f}}-v_{2 f}$

Relative velocity of approach $=v_{1 i}-\left(-v_{2 i}\right)$

$e=\frac{- \text { (relative velocity of separation) }}{\text { (relative velocity of approach) }}$

$e=\frac{-(v_{1 f}-v_{2 f})}{(v_{1 i}+v_{2 i})}$

$e\left(v_{1 i}+v_{2 i}\right)=v_{2 f}-v_{1 f}$

$m_1 v_{1 i}-m_2 v_{2 i}=m_1 v_{1 f}+m_2 v_{2 f}$ ...(1)

$e=\frac{v_{2 f}-v_{1 f}}{v_{2 i}+v_{1 i}}$

$v_{2 f}=v_{1 f}+e(v_{2 i}+v_{1 i})$ ...(2)

Substitution in (1)

$m_1 v_{1 i}-m_2 v_{2 i}=m_1 v_{1 f}+m_2(v_{1 f}+e v_{2 i}+e v_{1 i})$

$e\left(v_{1 i}+v_{2 i}\right)=v_{2 f}-v_{1 f}$

$v_{1 f}=\frac{m_1 v_{1 i}-m_2\left(v_{2 i}+e v_{2 i}+e v_{1 i}\right)}{\left(m_1+m_2\right)}$

$v_{2 f}=\frac{m_1\left(v_1 i+e v_1 i+e v_{2 i}\right)-m_2 v_{2 i}}{m_1+m_2}$

$m_1=m_2$

Elastic collision $e=1$

Platic collision or totally inelastic collision $e=0$

$m_1 v_{1 i}-m_2 v_{2 i}=\left(m_1+m_2\right) v_{2 f}$

$e=1$, corresponds to

$\frac{1}{2} m_1 v_{1 i}^2+\frac{1}{2} m_2 v_{2 i}^2=\frac{1}{2} m_1 v_{1 f}^2+\frac{1}{2} m_2 v_{2 f}^2$

(for the head on collision)

Head on collision

Analysis is done assuming $v_{2 i}=0$

Even if $v_{2 i} \neq 0$,

We can do a change of reference frames and study the motion $x$ a frame moving with a constant velocity of $v_{2 i}$.

In this frame $v_{2 i}=0$

$(\vec{v_{1 i}})$= $\vec{v_{1 i}}$= -$\vec{v_{2 i}}$

Frame where $\overrightarrow{v_{2 i}}=0$

Oblique elastic collision between 2 bodies of equal masses.

If we write the equation of conservation of energy:

$\frac{1}{2} m_1 v_{1 i}^2=\frac{1}{2} m_1 v_{1 f}^2+\frac{1}{2} m_2 v_{2 f}^2$

$v_{1 i}^2=v_{1 f}^2+v_{2 f}^2$

Three form right angle triangle.

$v_{i f}$ has to be perpendicular to $v_{2 f}$.

Body $m_2$ is very massive

$m_2 \gg m_1$

$m_2$ could be earth.

$v_{1 f}=\frac{m_1 v_{1 i}-m_2\left(v_{2 i}+e v_{2 i}+e v_{1 i}\right)}{\left(m_1+m_2\right)}$

Divide RHS (Numerator & Denominator by $\mathrm{m}_2$ )

$v_{i f}=\frac{\frac{m_1}{m_2} v_{1 i}-\left(v_{2 i}+e v_{2 i}+e v_{1 i}\right)}{\left(\frac{m 1}{m_2}+1\right) If v_{2 i}=0}$.

$v_{2 i}=0$

$v_{1 f}=-e v_{1 i}$

$v_{2 f}=-v_{2 i}$