Inelastic Collision in One Dimension

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An inelastic collision in one dimension is defined by some loss of kinetic energy during the event. Most everyday collisions result in some kinetic energy loss, making them inelastic by nature.

If colliding bodies remain joined post-collision, such as mud sticking to a wall after being thrown, the collision is termed perfectly inelastic. This is also true for arrows that embed in targets and meteorites that strike the Earth, integrating into its mass.

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Let’s suppose that there are two bodies of masses $ \displaystyle {{m}_{1}}$ and $ \displaystyle {{m}_{2}}$ and they are moving with velocities of $ \displaystyle {{u}_{1}}$ and $ \displaystyle {{u}_{2}}$ respectively along the same axis. They have a collision with each other and losses some of the kinetic energy. Therefore, the collision is inelastic as there is some loss of kinetic energy.

Consider that after the collision $ \displaystyle {{v}_{1}}$ and $ \displaystyle {{v}_{2}}$ are their velocities.

Inelastic Collision in One Dimension — Image Credit: © Briligence.com

As the two bodies form one system, which is closed and isolated, we can write the law of conservation of linear momentum for the two body system as:

Total momentum before the collision ($ \displaystyle {{P}_{i}}$) = Total momentum after the collision ($ \displaystyle {{P}_{f}}$)

$ \displaystyle {{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}$ $ \displaystyle ={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\ldots (1)$

The overhead arrows for vectors have been avoided as the collision is one dimensional and velocity components along one axis are used.

If we know masses $ m_1, m_2 $, initial velocities $ u_1, u_2 $ and one of the final velocities, we can calculate the other final velocity from eqn. (1).

Perfectly inelastic collision in one dimension

The figure below shows perfectly inelastic collision between two bodies of masses $ m_1 $ and $ m_2 $. the body of mass $ m_2 $ happens to be initially at rest ($ u_2 = 0 $). We refer to this body as the target. The incoming body of mass $ m_1 $, moving with initial velocity $ u_1 $ is referred to as the projectile. After the collision, the two bodies move together with a common velocity $ V $. The collision is perfectly inelastic.

As the total linear momentum of the system cannot change, therefore

\[
P_i = P_f
\]

i.e. \[
m_1 u_1 + m_2 u_2 = (m_1 + m_2) V
\]

$ \displaystyle {{m}_{1}}{{u}_{1}}=({{m}_{1}}+{{m}_{2}})V$ $ \displaystyle \ldots (\because \text{ }{{u}_{2}}=0)$

$ \displaystyle V=\frac{{{{m}_{1}}{{u}_{1}}}}{{{{m}_{1}}+{{m}_{2}}}}\ldots (2)$

Knowing $ m_1, m_2 $ and $ u_1 $, we can calculate the final velocity $ V $. As the mass ratio

$ \displaystyle \frac{{{{m}_{1}}}}{{{{m}_{1}}+{{m}_{2}}}}<1,$ $ \displaystyle \text{therefore, }V<{{u}_{1}}$

We can calculate loss of K.E. in this collision.

Total K.E. before collision = $ E_1 = \frac{1}{2} m_1 u_1^2 $

Total K.E. after collision = $ E_2 = \frac{1}{2} (m_1 + m_2) V^2 $

$ \displaystyle =\frac{1}{2}({{m}_{1}}+{{m}_{2}}){{\left( {\frac{{{{m}_{1}}{{u}_{1}}}}{{{{m}_{1}}+{{m}_{2}}}}} \right)}^{2}}$ $ \displaystyle \quad \text{using }(2)$

\[
= \frac{m_1^2 u_1^2}{2 (m_1 + m_2)}
\]

Loss of K.E. $ = E_1 – E_2 $

\[
= \frac{1}{2} m_1 u_1^2 – \frac{m_1^2 u_1^2}{2 (m_1 + m_2)}
\]

\[
= \frac{m_1^2 u_1^2 + m_1 m_2 u_1^2 – m_1^2 u_1^2}{2 (m_1 + m_2)}
\]

\[
= \frac{m_1 m_2 u_1^2}{2 (m_1 + m_2)} \quad (3)
\]

which is positive.

Remember: In a closed isolated system containing a collision, the linear momentum of each colliding body may change, but the total linear momentum $ P $ of the system remains constant, whether the collision is elastic or inelastic.

Inelastic Collision in Two Dimensions

In cases where two objects initially travel along the same straight path but, after the collision, move in different directions and lose some kinetic energy, the collision is termed an inelastic collision in two dimensions.

For such a system, where no external forces are involved, the total linear momentum is conserved. Therefore,

\[
\vec{P}_f = \vec{P}_i
\]

From above figure, the momentum components in the X-direction after the collision ($ \displaystyle {{P}_{f}}$) should equal those before the collision ($ \displaystyle {{P}_{i}}$):

$ \displaystyle {{m}_{1}}{{v}_{1}}\cos \theta +{{m}_{2}}{{v}_{2}}\cos \phi $ $ \displaystyle ={{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}\ldots (4)$

Since the total initial Y-axis momentum is zero, applying the conservation of linear momentum along the Y-axis gives:

$ \displaystyle 0={{m}_{1}}{{v}_{1}}\sin \theta -{{m}_{2}}{{v}_{2}}\sin \phi \ldots (5)$

Using these equations, along with the values for $m_1$, $m_2$, $u_1$, $u_2$, $\theta$, and $\phi$, we can determine the final velocities $v_1$ and $v_2$.

Perfectly Inelastic Collision in Two Dimensions

In the figure below, a perfectly inelastic collision is illustrated between two bodies, one with mass $m_1$ and the other with mass $m_2$. Initially, body $m_2$ moves along the X-axis with velocity $u_2$, while body $m_1$ is moving at an angle $\theta$ with respect to the X-axis with velocity $u_1$.

collision at O, the two bodies stick to each other and move with a common velocity V at an angle $\theta$ with X-axis as shown in the figure.

As the system is closed and isolated, the total linear momentum of the system remains constant.

Referring to the above figure, and equating initial momentum along X-axis to final momentum along the same axis, we get

$\displaystyle {{m}_{1}}{{u}_{1}}\cos \theta +{{m}_{2}}{{u}_{2}}$ $ \displaystyle =({{m}_{1}}+{{m}_{2}})V\cos \phi $

Again, applying the law of conservation of linear momentum along Y-axis, we get

$\displaystyle {{m}_{1}}{{u}_{1}}\sin \theta +0$ $ \displaystyle =({{m}_{1}}+{{m}_{2}})V\sin \phi \ldots (6)$

Knowing $m_1$, $m_2$, $u_1$, $u_2$, and $\theta$, we can calculate final velocity V and its direction i.e., $\phi$ from eqns. (5) and (6).

Example: A ball moving with a speed of 9 m/s strikes an identical ball at rest, which after the collision, the direction of each ball makes an angle of $30^\circ$ with the original line of motion. Find the speeds of the two balls after the collision.

Solution:

Here, $m_1 = m_2 = m$,
$u_1 = 9$ m/s, $u_2 = 0$,
$\theta_1 = \theta_2 = 30^\circ$, $v_1 = v_2 = ?$

Applying principle of conservation of linear momentum

(a) Along the direction of motion (X-axis)

$ \displaystyle {{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}$ $\displaystyle ={{m}_{1}}{{v}_{1}}\cos {{\theta }_{1}}+{{m}_{2}}{{v}_{2}}\cos {{\theta }_{2}}$

$ \displaystyle m\times 9+0$ $\displaystyle =m\times {{v}_{1}}\cos {{30}^{{}^\circ }}+m\times {{v}_{2}}\cos {{30}^{{}^\circ }}$

$ \displaystyle 9={{v}_{1}}\sqrt{3}/2+{{v}_{2}}\sqrt{3}/2$ $\displaystyle =({{v}_{1}}+{{v}_{2}})\sqrt{3}/2$

$ \displaystyle {{v}_{1}}+{{v}_{2}}=\frac{{18}}{{\sqrt{3}}}\ldots (i)$

(b) Along a direction perpendicular to the direction of motion (Y-axis)

$ \displaystyle 0+0$ $\displaystyle ={{m}_{1}}{{v}_{1}}\sin {{\theta }_{1}}-{{m}_{2}}{{v}_{2}}\sin {{\theta }_{2}}$

$$
m_1 v_1 \sin 30^\circ = m_2 v_2 \sin 30^\circ
$$

$$
m v_1 \sin 30^\circ = m v_2 \sin 30^\circ
$$

$$
v_1 = v_2
$$

Put in (i):

$$
2v_1 = \frac{18}{\sqrt{3}}
$$

$$
v_1 = \frac{9}{\sqrt{3}} = 3 \sqrt{3} \text{ m/s}
$$

Hence the two balls move with the same velocity $v = 3 \sqrt{3}$ m/s after collision.