WORK, ENERGY AND POWER-Notes
Physics - Notes
Scalar Product
In physics, many situations involve two vectors acting together, but the final result we care about is often
just a number (a scalar). The scalar product, also called the dot product, is a mathematical tool that does
exactly this—it combines two vectors and produces a scalar.
In Work, Power, and Energy, this idea becomes especially important because work done by a force is defined
using the scalar product.
Meaning of Scalar Product
- the magnitudes of the vectors, and
- the angle between them.
- If \(\vec{A}\) and \(\vec{B}\) are two vectors making an angle \(\theta\) with
each other, their scalar product is defined as:
\[\boxed{\;
\vec{A}\cdot\vec{B}=AB\, \cos \theta\;}
\]
where,
\(A=\mid A\mid\) is the magnitude of vector \(\vec{A}\)
\(B=\mid B\mid\) is the magnitude of vector \(\vec{B}\)
\(\theta\) is the smaller angle between the two vectors.
The result is a scalar, not a vector. It has magnitude only, no direction. - Scalar product follows the commutative law : \[\vec{A}\cdot\vec{B}=\vec{B}\cdot\vec{A}\]
- Scalar product obeys the distributive law: \[\vec{A}\cdot(\vec{B}+\vec{C})=\vec{A}\cdot\vec{B}+\vec{A}\cdot\vec{C}\]
- Further, \[\vec{A}\cdot(\lambda \vec{A}\cdot\vec{B})=\lambda(\vec{A}\cdot\vec{B})\]
-
For unit vectors \(\hat{i}\,\hat{j}\,\hat{k}\)
\[\hat{i}\cdot\hat{i}=\hat{j}\cdot\hat{j}=\hat{k}\cdot\hat{k}=1\] \[\hat{i}\cdot\hat{j}=\hat{j}\cdot\hat{k}=\hat{j}\cdot\hat{k}=0\] - Scalar Product in Terms of Components \[\begin{aligned}\vec{A}=A_x\,\hat{i}+A_y\,\hat{j}+A_y\,\hat{k}\\ \vec{B}=B_x\,\hat{i}+B_y\,\hat{j}+B_y\,\hat{k}\end{aligned}\] then their scalar product is: \[\vec{A}\cdot\vec{B}=A_xB_x+A_yB_y+A_zB_z\]
The scalar product of two vectors depends on:
Example-1
Find the angle between force \(\mathrm{F} = (3\hat{i} +4\hat{j}-5\hat{k} )\) kunit and displacement \(\mathrm{d} = (5\hat{i} +4\hat{j} +3\hat{k})\) unit. Also find the projection of F on d.
Solution
$$\begin{aligned}F&=3\hat{i}+4\hat{j}-5\hat{k}\\ d&=5\hat{i}+4\hat{j}+3\hat{k}\\\\ F\cdot d&=Fd\,\cos \theta \\\\ F\cdot d&=\left( 3\hat{i}+4\hat{j}-5\hat{k}\right)\cdot \left( 5\hat{i}+4\hat{j}+3\hat{k}\right) \\ &=15\,\hat{i}\cdot \hat{i}+12\,\hat{i}\cdot \hat{j}+9\,\hat{i}\cdot \hat{k}+20\,\hat{j}\cdot \hat{i}+16\,\hat{j}\cdot \hat{j}+12\,\hat{j}\cdot \hat{k}-25\,\hat{k}\cdot \hat{i}-20\hat{k}\cdot \hat{j}-15\hat{k}\cdot \hat{k}\\ &=15+16-15\\ &=16\\\\ \mid F\mid &=\sqrt{3^{2}+4^{2}+\left( -5\right) ^{2}}\\ &=\sqrt{9+16+25}\\ &=\sqrt{50}\\\\ \mid d\mid &=\sqrt{5^{2}+4^{2}+3^{2}}\\ &=\sqrt{25+16+9}\\ &=\sqrt{50}\\\\ F\cdot d&=FD\,\cos\theta \\ 16&=\sqrt{50}\sqrt{50}\cos \theta \\ \cos \theta &=\dfrac{16}{50}\\ \cos \theta &=\dfrac{8}{25}\\ \theta &=\cos ^{-1}\left( \dfrac{8}{25}\right) \end{aligned}$$THE WORK-ENERGY THEOREM
The Work–Energy Theorem is one of the most powerful and elegant results in mechanics. It provides a direct connection between force, motion, and energy, allowing us to analyze motion without tracking every detail of acceleration and time.
Relation for rectilinear motion under constant acceleration \(a\) is given by \[ v^2-u^2=2as \]
where \(u\) and \(v\) are the initial and final speeds and \(s\) the distance traversed. Multiplying both sides by \(m/2\), we have \[\begin{align} \dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2&=mas\tag{1}\\ &=Fs \end{align}\]
We can generalise Eq. (1) to three dimensions by employing vectors
\[v^2-u^2=2\,\vec{a}\cdot\vec{d}\]
Here \(a\) and \(d\) are acceleration and displacement
vectors of the object respectively.
Once again multiplying both sides by \(m/2\) , we obtain
\[\begin{aligned}\dfrac{1}{2}m\,v^2-\dfrac{1}{2}m\,u^2&=m\,\vec{a}\cdot\vec{d}\\
&=\vec{F}\cdot\vec{d}\end{aligned}\]
which implies
\[K_{final}-K_{initial}=W\]
The change in kinetic energy of a particle is equal to the work done on it by the net force.
Example-2
It is well known that a raindrop falls under the influence of the downward gravitational force and the opposing resistive force. The latter is known to be proportional to the speed of the drop but is otherwise undetermined. Consider a drop of mass 1.00 g falling from a height 1.00 km. It hits the ground with a speed of 50.0 m s-1. (a) What is the work done by the gravitational force ? What is the work done by the unknown resistive force?
Solution
Mass of the drop = 1 gm
Height from drop is falling = 1km = 1000m
velocity at the ground = 50 \(\mathrm{m\ s^{-1}}\)
Kinetic Energy of the drop
Work done by The gravitational force
$$\begin{aligned}W_{g}&=mgh\\ &=10^{-3}\times 10\times 10^{3}\\ &=10\ J\end{aligned}$$ From work-energy theorem $$\Delta K=W_{g}+W_{r}$$where \(W_r\) is work done by resistive force
$$\begin{aligned}1.25&=10+w_{r}\\ \Rightarrow W_{r}&=1.25-10\\ &=-8.75\ J\end{aligned}$$WORK
The work done by the force is defined to be the product of component of the force in the direction of the displacement and the magnitude of this displacement.
Thus, \[W = (F \cos \theta )\ d = \vec{F}\cdot\vec{d}\]No work is done if :
- The displacement is zero
- The force is zero
- The force and displacement are mutually perpendicular.
- Work can be both positive and negative depending angle between Force and displacement \[\begin{aligned}\text{positive}&=0\le \theta \le 90^\circ\\ \text{negative}&=90^\circ \lt \theta \le 180\end{aligned} \]
- Work and Energy have the same dimensions, \([ML^2T^{–2}]\)
- The SI unit of these is joule (J)
Alternative Units of Work/Energy in J
| erg | \(\mathrm{10^{-7}\ J}\) |
| electron volt (eV) | \(\mathrm{1.6\times 10^{-19}\ J}\) |
| calorie (cal) | \(\mathrm{4.186 J}\) |
| kilowatt hour (kWh) | \(\mathrm{3.6\times 10^6 J}\) |
KINETIC ENERGY
if an object of mass \(\mathrm{m}\) has velocity \(v\), its kinetic energy \(\mathrm{K}\) is \[\mathrm{K}=\dfrac{1}{2}m\ \vec{v}\cdot\vec{v}=\dfrac{1}{2}mv^2\] Kinetic energy is a scalar quantity
The kinetic energy of an object is a measure of the work an object can do by the virtue of its motion.
Example-2
In a ballistics demonstration a police officer fires a bullet of mass 50.0 g with speed 200 m s-1 (see Table 5.2) on soft plywood of thickness 2.00 cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emergent speed of the bullet ?
Solution
Mass of the bullet = 50 gm
Speed of bullet \(v_i= 200\ m\ s^{-1}\)
Initial Kinetic Energy
Final kinetic Energy = 10%
$$\begin{aligned}K_{f}&=\dfrac{1000\times 10}{100}\\ &=100J\end{aligned}$$Emergent speed \(v_f\)
$$\begin{aligned}\dfrac{1}{2}mv_{f}^{2}&=100\\ v^{2}_f&=\dfrac{2\times 100}{m}\\ &=\dfrac{2\times 100}{50\times 10^{-3}}\\ &=\dfrac{2\times 100\times 1000}{50}\\ &=4000\\\\ v_{f}&=\sqrt{400\times 10}\\ &=20\sqrt{10}\\ &=63.2\ \mathrm{m\ s^{-1}}\end{aligned}$$WORK DONE BY A VARIABLE FORCE
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In many real physical situations, the force acting on a body does not remain constant. It may change with position, direction, or time. Such a force is called a variable force.
If the displacement \(\Delta x\) is small, we can take the force \(\Delta F(x)\) as approximately constant and the work done is then, work done is \[\Delta W = \Delta F(x)\ \Delta x\]
Adding successive rectangular areas in Fig. 5.3(a) we get the total work done as
\[W\cong \sum_{x_i}^{x_f} F(x)\ \Delta x\]where the summation is from the initial position \(x_i\)to the final position \(x_f\).
If the displacements are allowed to approach zero, then the number of terms in the sum increases without limit, but the sum approaches a definite value equal to the area under the curve in Fig. 5.3(b). Then the work done is
\[ \begin{aligned} W&=\lim_{\Delta x \to 0} \sum_{x_i}^{x_f} F(x)\ \Delta x\\\\ &=\int\limits_{x_i}^{x_f}F(x)\ dx \end{aligned} \]THE WORK-ENERGY THEOREM FOR A VARIABLE FORCE
The time rate of change of kinetic energy is
\[ \begin{aligned} K&=\dfrac{1}{2}m\ v^2\\ \dfrac{dK}{dt}&=\dfrac{d}{dt}\left(\dfrac{1}{2}m\ v^2\right)\\ &=m\dfrac{d v}{dt}v\\ &=ma\ v\\ &=F\ \dfrac{dx}{dt} \end{aligned} \]Thus
\[ dK=Fdx \]Integrating from the initial position \((x_i)\) to final position \((x_f )\), we have
\[ \begin{aligned} \int\limits_{K_i}^{K_f}dK&=\int\limits_{x_i}^{x_f}F\ dx\\\\ K_f-K_i&=\int\limits_{x_i}^{x_f}F\ dx \end{aligned} \] \[\boxed{\;K_f-K_i=W\;}\]THE CONCEPT OF POTENTIAL ENERGY
-
What Is Potential Energy?
Potential energy is the energy possessed by a body due to its position or its state of configuration in a force field. A body may be at rest and yet store energy, which becomes evident when it is allowed to move or undergo a change in shape.
-
Origin of Potential Energy
Potential energy originates from the work done against a force. When work is done against a conservative force, the supplied energy does not vanish; instead, it gets stored in the system as potential energy that can later perform work.
-
Gravitational Potential Energy
When a body of mass \(m\) is raised through a height \(h\) against gravity near the Earth’s surface, the work done is stored as gravitational potential energy.
\[ U = mgh \]
Here, \(g\) represents the acceleration due to gravity. This relation is valid when the variation of \(g\) with height is negligible.
-
Reference Level and Relativity of Potential Energy
Potential energy is relative and depends on the choice of reference level. The zero level of potential energy can be chosen arbitrarily. Only differences in potential energy are physically meaningful.
-
Elastic Potential Energy
When a spring is stretched or compressed, work is done against its restoring force. This work is stored as elastic potential energy. For a spring obeying Hooke’s law:
\[ U = \frac{1}{2}kx^2 \]
where \(k\) is the spring constant and \(x\) is the extension or compression of the spring.
-
Potential Energy and Conservative Forces
Potential energy can be defined only for conservative forces. A force is said to be conservative if the work done by it depends solely on the initial and final positions and not on the path taken. Gravitational and elastic forces are typical examples.
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Interconversion of Potential and Kinetic Energy
Potential energy can transform into kinetic energy and vice versa. A falling object converts gravitational potential energy into kinetic energy, while a stretched spring converts elastic potential energy into motion. When only conservative forces act, the total mechanical energy of the system remains constant.
THE CONSERVATION OF MECHANICAL ENERGY
One of the most fundamental ideas in mechanics is that energy is not destroyed; it only changes its form. The principle of conservation of mechanical energy formalizes this idea for systems where only certain types of forces are involved and provides a powerful way to analyze motion without dealing directly with forces at every point.
-
1. Meaning of Mechanical Energy
Mechanical energy of a body or a system is the sum of:
- Kinetic energy (energy due to motion).
- Potential energy (energy due to position or configuration).
In compact form, the total mechanical energy is written as
\[\boxed{\; E_{\text{mechanical}} = K + U \;}\]
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2. Statement of the Principle
When only conservative forces act on a system, the total mechanical energy of the system remains constant.
Mathematically,
\[\boxed{\; K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}} \;}\]
Although kinetic and potential energies may change individually, their sum remains unchanged.
-
3. Physical Explanation
Conservative forces, such as gravity and elastic spring force, have the special property that:
- The work done by them depends only on the initial and final positions.
- Energy lost in one form appears entirely in another form.
When a body moves under such forces:
- A decrease in potential energy leads to an equal increase in kinetic energy.
- An increase in potential energy results in an equal decrease in kinetic energy.
There is no net loss of mechanical energy in the system.
-
4. Illustration: A Falling Body
Consider a body released from rest at a height \(h\).
- At the top: \[\boxed{\; K = 0, \quad U = mgh \;}\]
- During the fall: Potential energy decreases and kinetic energy increases.
- Just before reaching the ground: \[\boxed{\; K = mgh, \quad U = 0 \;}\]
At every point during the motion, the total mechanical energy is
\[\boxed{\; K + U = mgh = \text{constant} \;}\]
-
5. Illustration: Spring–Mass System
When a spring is compressed or stretched:
- The kinetic energy of the moving mass is gradually converted into elastic potential energy.
- At maximum compression or extension, kinetic energy becomes zero.
Throughout the motion, as long as no energy is lost to friction,
\[\boxed{\; K + U = \text{constant} \;}\]
-
6. Conditions for Validity
The conservation of mechanical energy holds only when:
- Only conservative forces act on the system.
- Non-conservative forces like friction or air resistance are absent or negligible.
If non-conservative forces are present, part of mechanical energy is transformed into other forms such as heat or sound.
-
7. When Mechanical Energy Is Not Conserved
In the presence of friction:
- Mechanical energy decreases with time.
- Total energy is still conserved, but some energy appears as internal or thermal energy.
Thus, failure of mechanical energy conservation does not violate the law of conservation of energy; it only means energy has changed into non-mechanical forms.
-
8. Importance of the Principle
The conservation of mechanical energy:
- Simplifies problem-solving by avoiding force and acceleration calculations.
- Connects motion directly with energy changes.
- Helps predict speeds, heights, and displacements in various situations.
- Forms the basis for understanding oscillations, projectiles, and planetary motion.
Example-3
A bob of mass \(m\) is suspended by a light string of length L . It is imparted a horizontal
velocity \(v_o\) at the
lowest point A such that it completes a semi-circular trajectory in the vertical plane with the
string
becoming slack only on reaching the topmost point, C. This is shown in Fig. 5.6. Obtain an
expression for
(i) \(v_o\);
(ii) the speeds at pointsB and C;
(iii) the ratio of the kinetic energies(KB/KC) at B and C.
Comment on the nature of the trajectory of the bob after it reaches the point C.
Solution
At the lowest point \(A\), the bob is given a horizontal speed \(v_{0}\). The level of point \(A\) is taken as the reference where the gravitational potential energy is zero. Hence, the total mechanical energy at \(A\) is purely kinetic and is given by
\[ E = \frac{1}{2} m v_{0}^{2} \]
At this point, the forces acting on the bob are its weight \(mg\) acting vertically downward and the tension \(T_{A}\) in the string acting towards the centre. The net radial force provides the necessary centripetal force for circular motion, so
\[ T_{A} - mg = \frac{m v_{0}^{2}}{L} \]
At the highest point \(C\), the string is just slack, so the tension becomes zero. The only force providing centripetal acceleration is the weight of the bob. The total mechanical energy at \(C\) is the sum of kinetic energy and potential energy. Taking the vertical distance between \(A\) and \(C\) as \(2L\), the energy at \(C\) is
\[ E = \frac{1}{2} m v_{C}^{2} + 2mgL \]
At \(C\), the centripetal force condition with zero tension gives
\[ mg = \frac{m v_{C}^{2}}{L} \]
which simplifies to
\[ v_{C}^{2} = gL \quad \Rightarrow \quad v_{C} = \sqrt{gL} \]
Substituting \(v_{C}^{2} = gL\) in the energy expression at \(C\), the total mechanical energy becomes
\[ \begin{aligned} E &= \frac{1}{2} m gL + 2mgL\\\\ &= \frac{mgL + 4mgL}{2}\\\\ &= \frac{5}{2} mgL \end{aligned} \]
Since mechanical energy is conserved, the energy at \(A\) and \(C\) must be equal. Equating the total energy at \(A\) and \(C\),
\[ \frac{1}{2} m v_{0}^{2} = \frac{5}{2} mgL \] \[ v_{0}^{2} = 5gL \] \[ v_{0} = \sqrt{5gL} \]
This gives the required minimum speed at the lowest point for the bob to just reach the topmost point \(C\) with the string becoming slack there.
From the centripetal condition at \(C\),
\[ \begin{aligned} mg &= \frac{m v_{C}^{2}}{L}\\\\ \Rightarrow v_{C}^{2} &= gL\\\\ v_{C} &= \sqrt{gL} \end{aligned} \]
To find the speed at point \(B\), note that \(B\) is at a height \(L\) above the lowest point \(A\). Therefore, the potential energy at \(B\) is \(mgL\), and the kinetic energy is \(\frac{1}{2} m v_{B}^{2}\). The total mechanical energy at \(B\) is
\[ E = \frac{1}{2} m v_{B}^{2} + mgL \]
Using conservation of energy between \(A\) and \(B\), the total energy at \(B\) must equal the total energy at \(A\), which is \(\frac{1}{2} m v_{0}^{2} = \frac{5}{2} mgL\). Thus,
\[ \begin{aligned} \frac{1}{2} m v_{B}^{2} + mgL &= \frac{1}{2} m v_{0}^{2}\\\\ \frac{1}{2} m v_{B}^{2} + mgL &= \frac{1}{2} m \cdot 5gL\\\\ \frac{1}{2} \left( m v_{B}^{2} + 2mgL \right) &= \frac{5}{2} mgL\\\\ \frac{m}{2} \left( v_{B}^{2} + 2gL \right) &= \frac{5}{2} mgL\\\\ v_{B}^{2} + 2gL &= 5gL\\\\ v_{B}^{2} &= 3gL\\\\ v_{B} &= \sqrt{3gL} \end{aligned} \]
The kinetic energy at \(B\) is
\[ K_{B} = \frac{1}{2} m v_{B}^{2} = \frac{1}{2} m (3gL) = \frac{3}{2} mgL \]
The kinetic energy at \(C\) is
\[ K_{C} = \frac{1}{2} m v_{C}^{2} = \frac{1}{2} m (gL) = \frac{1}{2} mgL \]
Therefore, the ratio of the kinetic energies at \(B\) and \(C\) is
\[ \frac{K_{B}}{K_{C}} = \frac{\frac{3}{2} mgL}{\frac{1}{2} mgL} = 3 \]
After reaching point \(C\), the string becomes slack and can no longer provide the centripetal force required for circular motion. The bob then leaves the circular path and continues its motion as a projectile under the influence of gravity alone, following a parabolic trajectory instead of remaining on the circular arc.
THE POTENTIAL ENERGY OF A SPRING
A spring is a simple mechanical system that can store energy when it is deformed. This stored energy is known as the potential energy of a spring or elastic potential energy.
In an ideal spring, the spring force \(F_s\) is proportional to \(x\) where \(x\) is the displacement of the block from the equilibrium position. The displacement could be either positive [Fig. 5.7(b)] or negative [Fig. 5.7(c)]. This force law for the spring is called Hooke’s law and is mathematically stated as
\[F_s=-kx\]The constant k is called the spring constant. Its unit is \(N\ m^{-1}\). The spring is said to be stiff if k is large and soft if k is small.
Suppose that we pull the block outwards as in Fig. 5.7(b). If the extension is \((x)\)m, the work done by the spring force is
\[ \begin{aligned} W_s&=\int\limits_0^{x_m}F_s\ dx\\ &=-\int\limits_0^{x_m}kx\ dx\\ &=-\dfrac{1}{2}kx_m^2 \end{aligned} \]If the block is moved from an initial displacement \(x_i\) to a final displacement \(x_f\), the work done by the spring force \(W_s\) is
\[ \begin{aligned} W&=-\int\limits_{x_i}^{x_f}kx\ dx\\\\ &=-\dfrac{1}{2}\Bigl[kx^2\Bigr]_{x_i}^{x_f}\\\\ &=\dfrac{1}{2}kx_i^2 - \dfrac{1}{2}kx_f^2 \end{aligned} \]Thus the work done by the spring force depends only on the end points. Specifically, if the block is pulled from \(x_i\) and allowed to return to \(x_i\) ;
\[ \begin{aligned} W&=-\int\limits_{x_i}^{x_i}kx\ dx\\\\ &=-\dfrac{1}{2}\Bigl[kx^2\Bigr]_{x_i}^{x_i}\\\\ &=\dfrac{1}{2}kx_i^2 - \dfrac{1}{2}kx_i^2\\\\ &=0 \end{aligned} \]The work done by the spring force in a cyclic process is zero.
Thus, the spring force is a conservative force.
POWER
Power is defined as the time rate at which work is done or energy is transferred.
The average power of a force is defined as the ratio of the work, W, to the total time t taken
\[P_{avg}=\dfrac{W}{t}\]The work dW done by a force \(\vec{F}\) for a displacement dr is \(dW = \vec{F}\cdot d\vec{r}\). The instantaneous power can also be expressed as
\[\begin{aligned} P&=\vec{F}.\dfrac{f\vec{r}}{dt}\\\\ &=\vec{F}\cdot\vec{v} \end{aligned} \]where \(\vec{v}\) is the instantaneous velocity when the force is \(\vec{F}\).
Power, like work and energy, is a scalar quantity.
Its dimensions are [ML2T–3].
In the SI,its unit is called a watt (W). The watt is \(\mathrm{1\ J\ s^{–1}}\).
COLLISIONS
In mechanics, a collision is a short-duration interaction between two or more bodies during
which they exert strong forces on each other and exchange momentum.
For an isolated system of colliding bodies, the vector sum of their linear momenta remains
constant during the
collision, provided no external force acts on the system.
If two bodies of masses \(m_{1}\) and \(m_{2}\) move along a line with initial velocities
\(u_{1}\) and \(u_{2}\), and after collision
they move with velocities \(v_{1}\) and \(v_{2}\), then along that line the conservation of
momentum is written as \[ m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}. \] This
relation is valid for all types of collisions, whether elastic or inelastic, as long as the
system is free from external impulses.
Elastic Collision
Collisions are broadly classified by how kinetic energy behaves during the impact. In an elastic collision, both the total linear momentum and the total kinetic energy of the system remain unchanged between the instant just before and just after the collision. The colliding bodies separate after impact, and any deformation that occurs is fully reversible, so that there is effectively no permanent loss of kinetic energy to internal effects like heat or lasting distortion. Ideal elastic collisions are approximated in laboratory situations using low-friction setups, such as air tracks or smooth glancing collisions between hard spheres.
Inelastic Collision
In an inelastic collision, the total linear momentum of the system is still conserved, but the total kinetic energy after the collision is smaller than before. Part of the initial kinetic energy is transformed into other forms, such as heat, sound, or energy associated with permanent deformation of the bodies. Everyday collisions, like two vehicles colliding on a road or a clay ball hitting a wall and sticking, are typically inelastic because they involve noticeable loss of kinetic energy to such internal effects.
Special case of Collision
A particularly important special case is the perfectly inelastic collision, in which the colliding bodies stick together and move as a single combined mass after impact. This situation corresponds to the maximum possible loss of kinetic energy compatible with the conservation of momentum. For two bodies of masses \(m_{1}\) and \(m_{2}\) moving along a straight line with initial velocities \(u_{1}\) and \(u_{2}\), if they stick together and move with a common velocity \(v\) after collision, the conservation of momentum gives \[ m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v, \] from which the common velocity can be obtained.
Collisions in One Dimension
Consider first a completely inelastic collision in one dimension. Then, in Fig. 5.10, \[\theta_1=\theta_2=0\]
\[m_{1}v_{1i}=\left( m_{1}+m_{2}\right) v_{f}\] (momentum conservation) $$\begin{aligned}m_{1}v_{1i}&=\left( m_{1}+m_{2}\right) v_{f}\\ v_{f}&=\dfrac{m_{1}}{\left( m_{1}+m_{2}\right) }v_{1i}\end{aligned}$$The loss in kinetic energy in collision is
$$\begin{aligned}\Delta k&=\dfrac{1}{2}m_{1}v_{1i}^{2}-\dfrac{1}{2}\left( m_{1}+m_{2}\right) v_{f}^{2}\\ &=\dfrac{1}{2}m_{1}v_{1i}^{2}-\dfrac{1}{2}\left( m_{1}+m_{2}\right) \left[ \dfrac{m_{1}}{\left( m_{1}+m_{2}\right) }\cdot v_{1i}\right] ^{2}\\ &=\dfrac{1}{2}m_{1}v_{1i}^{2}-\dfrac{1}{2}\dfrac{m_{1}v_{1}^{2}i}{m_{1}+m_{2}}\\ &=\dfrac{1}{2}m_{1}v_{1i}^{2}\left[ 1-\dfrac{m_{1}}{m_{1}+m_{2}}\right] \\ &=\dfrac{1}{2}m_{1}v_{1i}^{2}\cdot \left[ \dfrac{m_{1}-m_{2}+m_{1}}{m_{1}+m_{2}}\right] \\ &=\dfrac{1}{2}m_{1}v_{1i}^{2}\left[ \dfrac{m_{2}}{m_{1}+m_{2}}\right] \\ &=\dfrac{1}{2}\dfrac{m_{1}m_{2}}{m_{1}+m_{2}}v_{1i}\end{aligned}$$Consider next an elastic collision then kinetic momentum and kinetic Energy conservation equation are
$$\begin{aligned}m_{1}v_{1i}&=m_{1}v_{1f}+m_{2}v_{2f}\\ m_{2}v_{2f}&=m_{1}v_{1i}-m_{1}v_{1f}\\ m_{2}&=\dfrac{m_{1}v_{1i}-m_{1}v_{1f}}{v_{2f}}\\ \dfrac{1}{2}m_{1}v_{1i}^{2}&=\dfrac{1}{2}m_{1}v_{1f}^{2}+\dfrac{1}{2}m_{2}v_{2f}^{2}\\ m_{1}v_{1i}^{2}-m_{1}v_{1f}^{2}&=\dfrac{m_{1}v_{1i}-m_{1}v_{1f}}{v_{2f}}\cdot v_{2f}^{2}\\ v_{1i}^{2}-v_{1f}^{2}&=\left( v_{1i}-v_{1f}\right) \cdot v_{2f}\\ v_{2f}&=\dfrac{v_{1i}^{2}-v_{1f}^{2}}{v_{1i}-v_{1f}}\\ v_{2f}&=v_{1i}+v_{1f}\end{aligned}$$ substituting this in equation-(i) $$\begin{aligned}m_{1}v_{1i}&=m_{1}v_{1f}+m_{2}\left( v_{1i}+v_{1f}\right) \\ m_{1}v_{1i}&=m_{1}v_{1f}+m_{2}v_{li}+m_{2}v_{1f}\\ m_{1}v_{1i}-m_{2}v_{1i}&=\left( m_{1}+m_{2}\right) v_{1}f\\ v_{1i}\left( m_{1}-m_{2}\right) &=\left( m_{1}+m_{2}\right) v_{1f}\\ v_{1f}&=\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}}v_{1}i\end{aligned}$$ and \[v_{2f}=\dfrac{2m_1v_{1i}}{m_1+m_2}\]- Case I :
If the two masses are equal
\[\begin{aligned}v_{1f}&=0\\ v_{2f}&=v_{1i} \end{aligned}\]The first mass comes to rest and pushes off the second mass with its initial speed on collision.
- Case II :
If one mass dominates, e.g. \(m_2 \gt\gt m_1\)
\[\begin{aligned}v_{1f}&\approx v_{1i}\\ v_{2f}&\approx 0 \end{aligned}\]The heavier mass is undisturbed while the lighter mass reverses its velocity.
Collisions in Two Dimensions
Fig. 5.10 also depicts the collision of a moving mass \(m_1\) with the stationary mass \(m_2\). Linear momentum is conserved in such a collision. Since momentum is a vector this implies three equations for the three directions {x, y, z}. Consider the plane determined by the final velocity directions of \(m_1\) and \(m_2\) and choose it to be the x-y plane. The conservation of the z-component of the linear momentum implies that the entire collision is in the x-y plane. The x- and y-component equations are
\[m_1v_{1i} = m_1v_{1f} \cos \theta_1 + m_2v_{2f} \cos \theta_2\] \[0 = m_1v_{1f} \sin \theta_1 − m_2v_{2f} \sin \theta_2\]If, further the collision is elastic,
\[\dfrac{1}{2}m_1v_{1i}^2=\dfrac{1}{2}m_1v_{1f}^2+\dfrac{1}{2}m_2v_{2f}^2\]
