Notes-Part-2-Class-11-Science-Physics-Chapter-4-Laws of Motion-Maharashtra Board

Proof : Change in kinetic energy is equal to work done by the conservative force.

Consider an object of mass m moving with velocity $\vec{u}$ experiencing a constant opposing force $\vec{F}$ which slows it down to $\vec{v}$ during displacement $\vec{s}$

As $\vec{u}$ and $\vec{F}$ are oppositely directed, the entire motion will be along the same line. In this case we need not use the vector form, just ± signs should be good enough.

If a = F/m  is the acceleration, we can write the relevant equation of motion as

v² − u² = 2 (−a)s …..(1) (negative acceleration for opposing force)

The work done by force is W = (ma).s = F.s

Multiplying throughout by m/2 to eq (1), we get

½ mu² – ½ mv² = (ma).s = F.s

Left-hand side is decrease in the kinetic energy and the right-hand side is the work done by the force. Thus, change in kinetic energy is equal to work done by the conservative force, which is in accordance with work-energy theorem.

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Examples :

(1) As a person jumps out of a stationary boat, the boat moves in the backward direction. Initially, the momentum of the system was zero because both of them were at rest. While jumping, the forward momentum of the person must be equal in magnitude to the backward momentum of the boat such that their sum is zero and the total linear momentum of the person and boat is conserved.

(2) When a bomb explodes in mid-air, the fragments fly off in different directions However, it is found that the vector sum of the momenta of all the fragments at any instant always remains equal to the initial momentum of the bomb in both magnitude and direction. Here, there is an increase in KE and production of heat, sound and light because the chemical PE of its explosive ingredients is partly converted to these forms.

(3). Rocket propulsion depends on the law of conservation of momentum as applied to a system consisting of the rocket and the exhaust gases. Most of a rocket on its launching pad is oxygen and fuel, all of which is eventually burnt to accelerate the small payload. Exhaust gases (combustion products) are ejected at very high speed from the nozzle at the tail end of the rocket engine. Because of the momentum of the ejected gases, the rocket receives a compensating momentum in the opposite direction to conserve the original momentum of the rocket-gas system.

Principle or law of conservation of linear momentum from Newton's laws of motion :

Consider the collision of two particles of masses m₁ and m₂, moving before the collision with constant velocities $\vec{u_1}$ and $\vec{u_2}$ in the laboratory frame of reference, see below Fig. After the collision they have velocities $\vec{v_1}$ and $\vec{v_2}$, again considered to be constant.

We shall assume that there are no external forces on the two particle system and the contact forces, which the particles exert on one another, act only over the short time interval Δt for which the particles are in contact during collision. The contact forces during collision act only between the particles of the system : they are internal forces of the system as distinguished from external forces which originate in bodies not belonging to the system.

Then, during the collision, particle 1 exerts a force F₂₁ (F_{on 2 by 1}) on particle 2. The impulse $\vec{F}_{21}$Δt  on particle 2 changes its linear momentum from  to

$\vec{F}_{21}$Δt = $m_2\vec{v_2} - m_2\vec{u_2}$  ……..(1)

(from Newton's second law of motion.)

By Newton's third law of motion, particle 2 simultaneously exerts a reaction force $\vec{F}_{12}$

($\vec{F}_{on\,1\,by\,2}$ ) on particle 1

$\vec{F}_{12}$ = $-\vec{F}_{21}$  ……(2)

The impulse $\vec{F}_{12}$Δt  on particle 1 changes its linear momentum from $m_1\vec{u_1}$ to $m_1\vec{v_1}$  :

$\vec{F}_{12}$Δt = $m_1\vec{v_1} - m_1\vec{u_1}$    ……..(2)

From Eqs. (1), (2) and (3),

$\vec{F}_{12}$Δt = $-\vec{F}_{21}$Δt

Or

$m_1\vec{v_1} - m_1\vec{u_1}$ = $-m_2\vec{v_2} - m_2\vec{u_2}$

Which on arranging gives

$m_1\vec{v_1} + m_2\vec{v_2}$ = $m_1\vec{u_1} + m_2\vec{u_2}$

∴ [Total momentum after collision] = [Total momentum before collision]

Thus, if two particles interact only with one another, i.e., there is no external force on the system of two particles, then their total linear momentum is conserved, which is the principle or law of conservation of linear momentum.

This principle holds good even when the number of particles involved is more than two.

One dimensional or head-on collision: A collision is said to be head-on if the colliding objects move along the same straight line, before and after the collision.

For a head-on collision in one dimension, if u₁ and u₂ are the velocities before collision, and v₁ and v₂ are the velocities after collision, then the relative velocity of body 1 with respect to body 2 before and after the collision are

(u₁ − u₂) and (v₁ − v₂), respectively.

. e = $-\frac{\text{relative velocity after collision}}{\text{relative velocity before collision}}$

= $-\frac{v_1-v_2}{u_1-u_2}$ = $\frac{v_2-v_1}{u_1-u_2}$

= $\frac{\text{velocity of separation}}{\text{velocity of approach}}$

In a head-on collision, the relative velocity of the two particles is reversed, so that e is defined in such a way as to be positive.

Explanation : In a head-on elastic collision, the velocity of separation is equal to the velocity of approach. In a perfectly inelastic collision, the velocity of separation is zero (v₁ = v₂).

Therefore, 1 > e ≥ 0 and e is a measure of the elasticity of the collision :

• If e = 1, the collision is elastic.
• If 1 > e ≥ 0, the collision is inelastic, with e = o for perfectly inelastic collision.

Head-on collisions between macroscopic solid objects are never elastic, and the total final kinetic energy is less than the total initial kinetic energy. The coefficient e is a measurable physical quantity and can be shown to be related to this loss of kinetic energy in a collision. The coefficient e (like coefficient of friction) depends on the materials of the bodies.

Coefficient of restitution during a head-on, elastic collision:

Consider the head-on collision of two particles of masses m₁ and m₂, moving before the collision with constant velocities u₁ and u₂ in the laboratory frame of reference (see Fig.).

After the collision they have velocities v₁ and v₂; again considered to be constant. Since the collision is head-on, i.e., the motion is confined to one dimension, the velocity vectors all lie in the same line and hence we can write the equations in scalar form with the usual sign convention. Therefore, by the principle of conservation of linear momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂  …..(1)

or m₁(u₁ − v₁) = m₂(v₂ − u₂)  …..(2)

Also, since the collision is elastic, total kinetic energy before collision = total kinetic energy after collision :

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

∴ m₁(u₁² − v₁²)=m₂(v₂² − u₂²)

Or m₁(u₁ − v₁) (u_{1 +} v₁)  = m₂(v₂ − u₂) (v₂ + u₂)    …….(3)

Dividing Eq. (3) by Eq. (2 ), we get,

u_{1 +} v₁ = v₂ + u₂      …….(4)

Rearranging the terms in Eq. (4),

u₁ − u₂ = v₂ − v₁   …….(5)

Here, (u₁ − u₂) is the velocity with which particle 1 approaches particle 2 before collision and (v₂ − v₁) is the velocity with which particle 2 separates away from particle 1 after collision.

Hence, in a head-on elastic collision, velocity of separation = velocity of approach.

From Eq. (5), the coefficient of restitution,

e = $\frac{v_2-v_1}{u_1-u_2}$ = 1

Particular cases:

Two particles of masses m₁ and m₂ move with initial velocities u₁ and u₂ such that particle 1 collides head-on elastically with particle 2.

Their respective velocities after the collision are

v₁ = $(\frac{m_1-m_2}{m_1+m_2})u_1+(\frac{2m_2}{m_1+m_2})u_2$

v₂ = $(\frac{2m_1}{m_1+m_2})u_1+(\frac{m_2-m_1}{m_1+m_2})u_2$

(i) If the bodies are of equal masses (or identical), m₁ = m₂

Substituting m₁ = m₂ in the above equations,

v₁ = $(0)u_1+(\frac{2m_2}{2m_2})u_2$ = u₂

and v₂ = $(\frac{2m_1}{2m_1})u_1+(0)u_2$ = u₁

This shows that in a head-on elastic collision of two particles of equal masses, the particles exchange their velocities.

Further, if the second particle is at rest before the collision, u₂ = 0, so that after the collision, v₁ = 0 and v₂  = u₁  that is, the first particle initially moving stops and the second one takes off with the initial velocity of the first.

(ii) If colliding body is much heavier and the struck body is initially at rest, i.e., m₁ >> m₂ and u₂= 0

Then,

v₁ ≅ $(\frac{m_1}{m_1})u_1$  ≅ u₁

and v₂ ≅ $(\frac{2m_1}{m_1})u_1$ ≅ 2u₁

that is, if a massive particle makes an elastic head-on collision with a light one at rest, the massive one continues its motion at almost the same speed, and the light one takes off at nearly twice this speed.

(iii) The body which is struck is much heavier than the colliding body and is initially at rest, i.e., m₁ << m₂ and u₂= 0

Then, ignoring m₁ in comparison to m₂, the final velocities are 

v₁ ≅ $(-\frac{m_1}{m_1})u_1$  ≅ −u₁ and v₂ ≅ (0)u₁  ≅ 0

Thus, if a light particle makes an elastic collision with a massive one at rest, it rebounds with almost its initial speed; the massive one is almost unaffected.

Loss in the kinetic energy :

Loss in K.E. = Δ (K.E.) = Total initial K.E. − Total final K.E.

= ½ (m₁+m₂)v² – (½ m₁u₁² + ½ m₂u₂²)

= ½ (m₁+m₂)$(\frac{m_1u_1+m_2u_2}{m_1+M-2})^2$ – (½ m₁u₁² + ½ m₂u₂²)

= $\frac{1}{2}[\frac{(m_1u_1+m_2u_2)^2}{m_1+m_2}-(m_1u_1^2+m_2u_2^2)]$

= $\frac{1}{2}[\frac{m_1u_1^2+2m_1m_2u_1u_2+m_2^2u_2^2-m_1^2u_1^2-m_2^2u_2^2-m_1m_2u_1^2-m_1m_2u_2^2}{m_1+m_2}]$

= $\frac{1}{2}[-\frac{m_1m_2(u_1^2+u_2^2-2u_1u_2)}{m_1+m_2}]$ = $-\frac{1}{2}(\frac{m_1m_2}{m_1+m_2})(u_1-u_2)^2$

Since the bracketed terms are always positive, the minus sign in Eq. (2) shows that the final kinetic energy is less than the initial kinetic energy. That is, there is a loss in kinetic energy.

Loss in KE = initial KE — final KE = $\frac{1}{2}(\frac{m_1m_2}{m_1+m_2})(u_1-u_2)^2$

Consider a one-dimensional (head-on) collision of two particles, of masses m₁ and m₂, moving with constant initial velocities u₁ and u₂. If the coefficient of restitution is e, it can be shown that

Loss in KE = $\frac{1}{2}(\frac{m_1m_2}{m_1+m_2})(u_1-u_2)^2(1-e^2)$

Since 0 ≤ e ≤ 1, (1 — e²) is either zero or positive.

The term (u₁ — u₂)² on the right hand side is always positive. Hence, for 0 <e < 1, i.e., an inelastic collision, there is always a loss in kinetic energy.

(i) For an elastic collision, e = 1, and loss in KE = 0.

That is, KE is conserved in an elastic collision.

(ii) For a perfectly inelastic collision, e = 0, and

Loss in KE = $\frac{1}{2}(\frac{m_1m_2}{m_1+m_2})(u_1-u_2)^2$ which is maximum.

Consider a head-on (one-dimensional) collision of two identical objects, moving with constant initial velocities u₁ and u₂. Since they are identical,

m₁ = m₂ = m

If v₁ and v₂ are their velocities after the collision, then by the principle of conservation of linear momentum, *

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

∴ u₁ + u₂ = v₁ + v₂

and the coefficient of restitution,

e = $\frac{v_2-v_1}{u_1-u_2}$ = $\frac{v_2-v_1}{u_1}$ ….Since the target was at rest, u₂ = 0.

∴ u₁ = v₁ + v₂     …..(1)

and v₂ – v₁= eu₁ ...(2)

∴ $\frac{1}{e}=\frac{v_1+v_2}{v_2-v_1}$

∴ $\frac{v_2}{v_1}=\frac{1+e}{1-e}$

∴ $\frac{v_1}{v_2}=\frac{1-e}{1+e}$   (0<e<1)

is the required expression.

Consider a head-on (one-dimensional) collision of two particles, of masses m₁ and m₂, moving with constant initial velocities u₁ and u₂. If v₁ and v₂ are their velocities after the collision, then by the principle of conservation of linear momentum,

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂   …. (1)

The coefficient of restitution for the inelastic collision,

0 < e = $\frac{v_2-v_1}{u_1-u_2}$ < 1

∴ v₂ = v₁ + eu₁ − eu₂ ...(2)

Substituting this value of v₂ in Eq. (1),

m₁u₁ + m₂u₂ = m₁v₁ + m₂(v₁ + eu₁ − eu₂)

= m₁v₁ + m₂v₁ + em₂u₁ − em₂u₂

∴ m₁u₁ - em₂u₁ + m₂u₂ + em₂u₂ = (m₁ + m₂)v₁

∴ v₁ = $(\frac{m_1-em_2}{m_1+m_2})u_1+(\frac{m_2(e+1)}{m_1+m_2})u_2$  ……. (3)

Similarly, from Eq. (2), v₁ = v₂ − eu₁ + eu₂, which on substitution into Eq. (1) gives

v₂ = $(\frac{m_1(1+e)}{m_1+m_2})u_1+(\frac{m_2-em_1}{m_1+m_2})u_2$  ……. (4)

Equations (3) and (4) give the required expressions.

Equations (3) and (4) can also be written as

v₁ = $\frac{em_2(u_2-u_1)+m_1u_1+m_2u_2}{m_1+m_2}$ and

v₂ = $\frac{em_1(u_1-u_2)+m_1u_1+m_2u_2}{m_1+m_2}$

The impulse of a force :

For a force $vec{F}$ which is constant over the time interval Δt during which it acts, the impulse of the force is defined as the product $\vec{F}$Δt :

$\vec{J}$ = $\vec{F}$Δt

Newton's second law of motion, which duplicates the first law with quantitative precision, states that the change in momentum of a body equals the impulse of the applied force and is made in the direction of that force.

Then,

$\vec{J}=\vec{dp}=\vec{p_f}-\vec{p_i}=m(\vec{v}-\vec{u})$

where m is the mass of the body $\vec{p_i}$ and $\vec{p_f}$ are the initial and final momenta of the body over the time interval of the impulse.

For a time-varying force $\vec{F}(t)$, whose magnitude over the interval Δt is given by the curve in Fig. (a), the impulse of the force is given by the area under the curve :

$\vec{J}=\int_{t_i}^{t_f}\vec{F}dt$

For a time-varying force, the magnitude of the impulse can also be written as

J = F_av Δt

where F_av is the average force that would impart the same impulse as the variable force over the time interval, i.e., the area F_av Δt of the rectangle in Fig. (b) is equal to the area under the curve in Fig. (a).

Dimensions : [Impulse] = [force]['dme]

= [mass] [acceleration] [time]

= [mass][vel0city]

= [MLT⁻¹]

SI unit : The newton-second, N.s (= kg.m/s)

Collision in two dimensions, i.e., a nonhead-on collision:

In this case, the direction of at least one initial velocity is not along the line of impact.

Consider an oblique impact/collision of two smooth spheres, moving with velocities $\vec{u_1}$ and $\vec{u_2}$ . At the moment of impact, the two spheres have a common tangent, which is perpendicular to the line through the centres of the two spheres.

The line through the centres of the two spheres is called the line of impact. The mutual forces between the two spheres act along the line of impact, so the impulse affecting each sphere also acts along the line of impact.

Let us consider the line of impact as the x-axis and the common tangent as the y-axis, see above Fig.

Then, according to the law of conservation of linear momentum along the line of impact,

m₁u_1x + m₂u_2x = m₁v_1x + m₂v_2x  

m₁u₁ cos α₁ + m₂u₂ cos α₂ = m₁v_1x cos β₁ + m₂v₂ cos β₂     …… (1)

The components of the velocities along the common tangent are unchanged in the impact.

u_1y = v_1y and u_2y = v_2y

∴ u₁ sin α₁ = v₁ sin β₁      ….(2)

and u₂ sin α₂ = v₂ sin β₂   ……(3)

Newton's law of restitution applies to the components of the velocities of the spheres parallel to the line of impact. Thus, the coefficient of restitution between the spheres is

e = $\frac{v_{2x}-v_{1x}}{u_{2x}-u_{1x}}$

  = $-\frac{v_2cosβ_2-v_1cosβ_1}{u_2cosα_2-u_1cosα_1}$

  = $\frac{v_2cosβ_2-v_1cosβ_1}{u_1cosα_1-u_2cosα_2}$    ….(4)

Equations (1), (2), (3) and (4) are solved for the four unknowns v₁, v₂, β₁ and β₂. The magnitude of the impulse on sphere 2 by 1 is

J₁₂ = μ (u₁ cos α₁ − u₂ cos α₂)(1+e) 

and the magnitude of the impulse on sphere 1 by 2 is

J₂₁ = μ (u₂ cos α₂ − u₁ cos α₁)(1+ e) 

where μ  = $\frac{m_1m_2}{m_1+m_2}$  is the reduced mass of the system.

The Loss in KE =  $\frac{1}{2}$ μ (u₁ − u₂)²(1 – e²)