Kinetics of Particle Notes FE Sem 1 Engineering Mechanics | MU

Kinetics relates the forces on a particle to the motion they produce. In the first part, D'Alembert's principle converts a dynamics problem into an equivalent statics problem by adding an inertia force, so the familiar equilibrium equations and free body diagrams can be used for lifts, blocks on planes and connected bodies. The second part uses the principle of linear impulse and momentum, which is especially useful when forces act for a short time, and applies conservation of momentum and the coefficient of restitution to direct and oblique central impact. The loss of kinetic energy in collisions of inelastic bodies completes the module.

Summary

Basic concepts of kinetics: mass is the quantity of matter and a measure of inertia; weight W = mg is the gravitational force on it. Newton's second law states that the net force on a particle equals the rate of change of its momentum, which for constant mass is F = ma, with the acceleration in the direction of the net force. Linear momentum is the product of mass and velocity, a vector measured in kg m/s or N s.

D'Alembert's principle and inertia force: rewriting F = ma as F + (-ma) = 0 shows that a moving particle can be treated as if in equilibrium under the real forces and an imaginary inertia force equal to ma acting opposite to the acceleration. This is D'Alembert's principle. The inertia force is not a real interaction force; it is a device that lets static methods be used.

Equations of dynamic equilibrium: with the inertia force included on the free body diagram, Sum Fx = 0 and Sum Fy = 0 apply (dynamic equilibrium). For motion along one direction the working equation becomes (forces in the direction of motion) - (resisting forces) - ma = 0. Typical applications are lifts, blocks on rough horizontal or inclined planes, and connected bodies over pulleys, where a separate FBD is drawn for each body and the string tension is common to both.

Principle of linear impulse and momentum: integrating F = m dv/dt over time gives the integral of F dt = m v2 - m v1. The left side is the impulse of the force, the product of force and time for a constant force (unit N s). The principle states that the impulse of the resultant force on a particle over an interval equals the change in its linear momentum in that interval. It is applied component-wise and is ideal for short impulsive forces such as a bat hitting a ball.

Law of conservation of momentum: if the resultant external impulse on a system of particles is zero, the total linear momentum of the system is conserved: m1 u1 + m2 u2 = m1 v1 + m2 v2. During collisions the impact forces are internal and very large compared with external forces such as weight, so momentum is conserved along the line of impact; examples include gun and bullet recoil and colliding bodies.

Coefficient of restitution: impact has a deformation period followed by a restitution period. The coefficient of restitution e is the ratio of the relative velocity of separation to the relative velocity of approach along the line of impact: e = (v2 - v1)/(u1 - u2). For perfectly elastic impact e = 1 and kinetic energy is conserved; for perfectly plastic impact e = 0 and the bodies move together; for real bodies 0 < e < 1. For a ball dropped on a fixed floor, e = square root of (h1/h) where h1 is the rebound height.

Direct central and oblique central impact: the line of impact is the common normal to the surfaces at contact. In central impact the mass centres lie on this line. In direct central impact the velocities are along the line of impact; momentum conservation and the restitution equation give the two final velocities. In oblique central impact one or both velocities are inclined to the line of impact: resolve them along and perpendicular to the line of impact. For smooth bodies, the perpendicular components remain unchanged, and the components along the line are treated as a direct impact.

Loss of kinetic energy in collision of inelastic bodies: in an impact with e < 1 some kinetic energy is converted into heat, sound and permanent deformation. The loss is Delta KE = [m1 m2 / (2(m1 + m2))] (1 - e^2)(u1 - u2)^2, where u1 - u2 is the relative approach velocity along the line of impact. It is zero for e = 1 and maximum for a perfectly plastic impact.

Mind map

Solving a direct or oblique central impact problem

Important definitions

Formulas

Key points to remember

• The inertia force acts opposite to the acceleration, not necessarily opposite to the velocity.
• Draw separate free body diagrams with inertia forces for each body in connected-body problems; tension is the same throughout a light inextensible string over a smooth pulley.
• Kinetic friction mu_k N opposes the relative motion and is used once the body moves.
• Impulse and momentum are vectors; always apply the principle along chosen axes with signs.
• Momentum is conserved in collisions, but kinetic energy is conserved only if e = 1.
• In oblique impact of smooth bodies, components perpendicular to the line of impact do not change.
• e is defined using velocity components along the line of impact only.
• In a plastic impact (e = 0) the bodies move together after impact with a common velocity.
• When a ball strikes a fixed smooth surface obliquely, the parallel component is unchanged and the normal component reverses and becomes e times its value.

Solved examples

Example 1. A lift of mass 500 kg is accelerated upward at 2 m/s^2. Find the tension in the cable. Also find the reading of a weighing machine on which a 60 kg man stands when the lift moves down with an acceleration of 1.5 m/s^2.

Upward acceleration: dynamic equilibrium T - mg - ma = 0, so T = m(g + a) = 500(9.81 + 2) = 5905 N. Man in lift moving down with a = 1.5 m/s^2: inertia force ma acts upward on him, so R + ma - mg = 0, R = m(g - a) = 60(9.81 - 1.5) = 498.6 N (about 50.8 kg reading).

Example 2. A 50 kg block is pulled up a 30 degree incline by a force of 400 N parallel to the plane. The coefficient of kinetic friction is 0.2. Using D'Alembert's principle find the acceleration.

Perpendicular to the plane: N = mg cos 30 = 50(9.81)(0.866) = 424.79 N. Friction (down the plane) = 0.2 x 424.79 = 84.96 N. Component of weight down the plane = mg sin 30 = 245.25 N. Along the plane, inertia force 50a acts down the plane: 400 - 245.25 - 84.96 - 50a = 0, a = 69.79/50 = 1.40 m/s^2 up the plane.

Example 3. A 20 kg block A on a rough horizontal table (mu_k = 0.25) is connected by a light inextensible string over a smooth pulley at the edge to a hanging block B of 15 kg. Find the acceleration and the string tension.

Block A (moving towards the pulley): T - 0.25(20)(9.81) - 20a = 0, so T = 49.05 + 20a. Block B (moving down): 15(9.81) - T - 15a = 0, so T = 147.15 - 15a. Equating: 49.05 + 20a = 147.15 - 15a, 35a = 98.1, a = 2.80 m/s^2. T = 147.15 - 15(2.80) = 105.1 N.

Example 4. A cricket ball of mass 0.5 kg moving at 20 m/s is hit back by a bat so that it returns along the same line at 30 m/s. If the contact lasts 0.01 s, find the impulse and the average force exerted by the bat.

Take the return direction as positive: initial velocity -20 m/s, final +30 m/s. Impulse = m(v2 - v1) = 0.5(30 - (-20)) = 25 N s. Average force = impulse/time = 25/0.01 = 2500 N in the return direction.

Example 5. A bullet of mass 30 g is fired horizontally at 400 m/s from a gun of mass 4 kg. Find the recoil velocity of the gun.

The system starts from rest and external horizontal impulse is negligible, so total momentum stays zero: 0.03(400) + 4 V = 0, V = -12/4 = -3 m/s. The gun recoils at 3 m/s opposite to the bullet.

Example 6. Two balls of masses 2 kg and 3 kg move towards each other along the same line with speeds 6 m/s and 4 m/s. If e = 0.6, find their velocities after impact and the loss of kinetic energy.

Take the 2 kg ball's direction as positive: u1 = 6, u2 = -4 m/s. Momentum: 2(6) + 3(-4) = 0 = 2v1 + 3v2. Restitution: v2 - v1 = 0.6(6 - (-4)) = 6. From the first equation v1 = -1.5 v2; so v2 + 1.5v2 = 6, v2 = 2.4 m/s and v1 = -3.6 m/s. Both balls reverse direction. KE before = 0.5(2)(36) + 0.5(3)(16) = 60 J. KE after = 0.5(2)(12.96) + 0.5(3)(5.76) = 21.6 J. Loss = 38.4 J. Check with formula: [2 x 3/(2 x 5)](1 - 0.36)(10^2) = 0.6 x 0.64 x 100 = 38.4 J.

Example 7. A ball strikes a smooth horizontal floor with a velocity of 10 m/s at 60 degrees to the horizontal. If e = 0.5, find its velocity after impact.

Line of impact is vertical. Horizontal component = 10 cos 60 = 5 m/s, unchanged (smooth floor). Vertical component before = 10 sin 60 = 8.66 m/s downward; after = e x 8.66 = 4.33 m/s upward. Rebound velocity = square root of (5^2 + 4.33^2) = 6.61 m/s at arctan(4.33/5) = 40.89 degrees to the floor.

Example 8. A smooth sphere A of mass 1 kg moving at 10 m/s strikes a stationary smooth sphere B of mass 2 kg of the same size. At impact the velocity of A makes 30 degrees with the line of impact. If e = 0.8, find the velocities of both spheres after impact and the loss of kinetic energy.

Along the line of impact (n): uA = 10 cos 30 = 8.66 m/s, uB = 0. Perpendicular (t): A has 10 sin 30 = 5 m/s, B has 0; these are unchanged. Momentum along n: 1(8.66) = vA + 2vB. Restitution: vB - vA = 0.8(8.66) = 6.93. Adding: 3vB = 15.59, vB = 5.20 m/s, vA = 8.66 - 10.39 = -1.73 m/s. A: components -1.73 (n) and 5 (t), speed = square root of (1.73^2 + 5^2) = 5.29 m/s at 109.1 degrees to the original line of impact direction. B: 5.20 m/s along the line of impact. KE before = 0.5(1)(100) = 50 J; after = 0.5(1)(5.29^2) + 0.5(2)(5.20^2) = 14 + 27 = 41 J; loss = 9 J. Check: [1 x 2/(2 x 3)](1 - 0.64)(8.66^2) = 0.3333 x 0.36 x 75 = 9 J.

Important questions and answers

Q1. State and explain D'Alembert's principle. (4 marks)

Newton's law F = ma can be written as F + (-ma) = 0. D'Alembert's principle states that the system of external forces acting on a body together with the inertia force (-ma) is in equilibrium. The inertia force has magnitude ma and acts opposite to the acceleration through the mass centre. Including it on the FBD converts a dynamic problem into a static one, so equations Sum Fx = 0 and Sum Fy = 0 can be used. Example: for a lift accelerating up, T - mg - ma = 0 gives T = m(g + a).

Q2. Define inertia force and explain dynamic equilibrium. (3 marks)

Inertia force is an imaginary force of magnitude equal to mass times acceleration, acting opposite to the acceleration. It represents the resistance of a body to change in its state of motion. When it is added to the real forces, the body is said to be in dynamic equilibrium, and the vector sum of all forces, real and inertia, is zero.

Q3. Derive the principle of linear impulse and momentum. (4 marks)

From Newton's second law F = m dv/dt, so F dt = m dv. Integrating between times t1 and t2 when velocities are v1 and v2: integral from t1 to t2 of F dt = m(v2 - v1). The left side is the impulse of the resultant force and the right side is the change in linear momentum. Hence the impulse of the resultant force on a particle equals the change in its momentum. It is applied component-wise and is useful when forces act over short times.

Q4. State the law of conservation of momentum and give two applications. (3 marks)

If the resultant external force (impulse) acting on a system is zero, the total linear momentum of the system remains constant: m1 u1 + m2 u2 = m1 v1 + m2 v2. Applications: recoil of a gun when a bullet is fired; collision of two bodies, where impact forces are internal; a person jumping from a boat; explosion of a shell into fragments.

Q5. Define coefficient of restitution and explain the types of impact based on it. (4 marks)

Coefficient of restitution e is the ratio of the relative velocity of separation to the relative velocity of approach along the line of impact, e = (v2 - v1)/(u1 - u2). Equivalently it is the ratio of the impulse during restitution to the impulse during deformation. Perfectly elastic impact: e = 1, no loss of KE. Perfectly plastic impact: e = 0, the bodies move together after impact and the loss of KE is maximum. Semi-elastic (real) impact: 0 < e < 1, partial loss of KE.

Q6. Differentiate between direct central impact and oblique central impact. (3 marks)

In both, the mass centres lie on the line of impact. In direct central impact the velocities of both bodies before impact are along the line of impact, so only one direction needs to be analysed. In oblique central impact the velocity of at least one body is inclined to the line of impact; velocities are resolved into components along and perpendicular to the line of impact. Along the line, momentum conservation and the restitution law apply; perpendicular to it, for smooth bodies, each body's component remains unchanged.

Q7. Derive the expression for loss of kinetic energy in direct central impact. (6 marks)

Let masses m1, m2 have velocities u1, u2 before and v1, v2 after impact. Momentum: m1(u1 - v1) = m2(v2 - u2) ... (1). Restitution: v2 - v1 = e(u1 - u2) ... (2). Loss of KE = (1/2)m1(u1^2 - v1^2) + (1/2)m2(u2^2 - v2^2) = (1/2)m1(u1 - v1)(u1 + v1) - (1/2)m2(v2 - u2)(v2 + u2). Using (1), with P = m1(u1 - v1) = m2(v2 - u2): loss = (P/2)[(u1 + v1) - (v2 + u2)] = (P/2)[(u1 - u2) - (v2 - v1)] = (P/2)(1 - e)(u1 - u2). Solving (1) and (2): v1 = u1 - m2(1 + e)(u1 - u2)/(m1 + m2), so P = m1 m2 (1 + e)(u1 - u2)/(m1 + m2). Hence loss of KE = [m1 m2/(2(m1 + m2))](1 - e^2)(u1 - u2)^2.

Q8. A ball dropped from a height of 9 m on a fixed floor rebounds to 2.25 m. Find e and the height of the second rebound. (3 marks)

Velocity of approach = square root of (2gh), velocity of separation = square root of (2g h1). e = square root of (h1/h) = square root of (2.25/9) = 0.5. Second rebound height h2 = e^2 h1 = 0.25 x 2.25 = 0.5625 m.

Q9. Two bodies of 10 kg and 5 kg moving in the same direction at 6 m/s and 3 m/s undergo a perfectly plastic impact. Find the common velocity and loss of KE. (3 marks)

Common velocity v = (10 x 6 + 5 x 3)/15 = 75/15 = 5 m/s. KE before = 0.5(10)(36) + 0.5(5)(9) = 180 + 22.5 = 202.5 J. KE after = 0.5(15)(25) = 187.5 J. Loss = 15 J. Check: [50/30](1 - 0)(3^2) = 15 J.

Exam tips

• Show the inertia force on the FBD with a dotted arrow opposite to the acceleration; marks are given for it.
• State the assumed direction of motion and check that friction opposes it.
• In impact problems fix a positive direction along the line of impact and keep signs for all velocities.
• Write both equations (momentum and restitution) explicitly before solving.
• For oblique impact, state that tangential components are unchanged because the surfaces are smooth.
• Verify loss of KE by the formula; it is a quick check of the final velocities.

FAQs

Is the inertia force a real force?

No. It is an imaginary force introduced by D'Alembert's principle so that a moving body can be analysed with equilibrium equations.

Is kinetic energy conserved in a collision?

Only in a perfectly elastic impact (e = 1). Momentum is conserved in every collision when external impulse is negligible.

Can e be greater than 1?

Not in ordinary collisions of passive bodies; e lies between 0 and 1. A value above 1 would require an energy source such as an explosion at contact.

Is work-energy principle part of this module?

The module content lists force and acceleration, D'Alembert's principle, impulse-momentum and impact. Work-energy appears in the prerequisite and course outcome, so know it, but focus on the listed topics.

These notes follow the University of Mumbai first-year engineering syllabus (NEP 2020, Engineering Mechanics) and were prepared with AI help and checked.