Dynamics studies the causes of motion: how forces change the motion of bodies. This unit builds on Newton's three laws to study momentum and impulse, the conservation of linear momentum, applications of Newton's laws, the turning effect of forces (torque) and equilibrium, and solid friction.
Newton's laws of motion (recap):
First law (inertia): A body stays at rest or in uniform motion in a straight line unless an external net force acts on it.
Second law: The rate of change of momentum of a body is proportional to the net force and takes place in the direction of the force: F = dp/dt; for constant mass, F = ma.
Third law: To every action there is an equal and opposite reaction. The two forces act on different bodies.
4.1 Linear Momentum and Impulse
4.1(a) Linear Momentum
The linear momentum of a body is the product of its mass and velocity:
p = m v
It is a vector in the direction of the velocity. SI unit: kg m s⁻¹ (equivalently N s); dimensions [MLT⁻¹]. A heavy lorry and a light car at the same speed have different momenta; the lorry is harder to stop.
Newton's second law in terms of momentum
F ∝ rate of change of momentum = (p₂ − p₁)/Δt
F = Δp / Δt = d(mv)/dt
For constant mass: F = m (dv/dt) = ma
4.1(b) Impulse
When a large force acts for a very short time (a bat striking a ball), the product of the force and the time of action is called impulse:
J = F Δt (constant force)
J = ∫ F dt (variable force)
Its SI unit is N s (the same as kg m s⁻¹), and it is a vector in the direction of the force.
Figure 4.1: Force–time graph of a collision. The area under the curve is the impulse. The dashed line is the average force Fav that would give the same impulse
Impulse–momentum theorem
From F = Δp/Δt:
F Δt = Δp = mv − mu
Impulse = change in momentum
The average force is Fav = Δp / Δt, so for a given change in momentum, a longer contact time means a smaller force.
Everyday applications:
A cricketer pulls the hands back while catching a fast ball: the time of stopping increases, so the force on the hands decreases.
Car airbags, padded helmets and crumple zones increase the stopping time.
A person jumping onto sand or a mattress is safer than jumping onto concrete.
Bending the knees on landing prolongs the stopping time.
Example 4.1
A cricket ball of mass 0.15 kg moving at 30 m s⁻¹ is hit back along the same line at 40 m s⁻¹. The contact time is 0.01 s. Find the impulse and the average force.
Take the final direction as positive: u = −30 m s⁻¹, v = +40 m s⁻¹
Δp = m(v − u) = 0.15 × (40 + 30) = 10.5 N s
Fav = Δp / Δt = 10.5 / 0.01 = 1050 N
Example 4.2
A constant force of 20 N acts for 0.5 s on a 2 kg body at rest. Find the impulse and the final speed.
J = F Δt = 20 × 0.5 = 10 N s
J = mv ⇒ v = 10 / 2 = 5 m s⁻¹
4.2 Conservation of Linear Momentum
Statement: If no external force acts on a system, the total linear momentum of the system remains constant.
Figure 4.2: Collision of two bodies on a smooth line: before, during and after. During the collision the forces are equal and opposite
Derivation from Newton's laws
Two bodies of masses m₁ and m₂ with velocities u₁ and u₂ collide. During contact, body 1 exerts a force F on body 2, and by Newton's third law body 2 exerts −F on body 1 for the same time Δt. After the collision the velocities are v₁ and v₂.
Impulse on body 1: −F Δt = m₁(v₁ − u₁)
Impulse on body 2: +F Δt = m₂(v₂ − u₂)
Adding: m₁(v₁ − u₁) + m₂(v₂ − u₂) = 0
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
The total momentum before the collision equals the total momentum after it.
4.2(a) Types of Collision
Type
Momentum
Kinetic energy
After collision
Elastic
Conserved
Conserved
Bodies separate
Inelastic
Conserved
Some lost (heat, sound, deformation)
Bodies separate
Perfectly inelastic
Conserved
Maximum loss
Bodies stick together
Momentum is conserved in every collision where no external force acts; kinetic energy is conserved only in elastic ones.
Elastic collision in one dimension
Conservation of momentum and of kinetic energy give
Special cases of an elastic collision (target at rest, u₂ = 0):
m₁ = m₂: v₁ = 0, v₂ = u₁. The velocities are exchanged.
m₁ ≫ m₂: v₁ ≈ u₁, v₂ ≈ 2u₁. The heavy body barely slows; the light body shoots ahead.
m₁ ≪ m₂: v₁ ≈ −u₁, v₂ ≈ 0. The light body bounces back; the heavy body stays nearly at rest.
4.2(b) Perfectly Inelastic Collision
m₁u₁ + m₂u₂ = (m₁ + m₂) v
v = (m₁u₁ + m₂u₂) / (m₁ + m₂)
4.2(c) Recoil of a Gun
Before firing, the gun and bullet are at rest, so the total momentum is zero. After firing, the bullet (mass m) moves forward with velocity v and the gun (mass M) recoils with velocity V.
Figure 4.3: Recoil of a gun. The total momentum stays zero
0 = mv + MV ⇒ V = − mv / M
The negative sign shows that the gun moves opposite to the bullet. The gun's recoil speed is small because its mass is large. The same principle explains the working of rockets: hot gas is expelled backward at high speed, and the rocket gains an equal forward momentum, so no air is needed for propulsion.
Example 4.3 — Gun recoil
A 5 kg gun fires a 0.02 kg bullet at 400 m s⁻¹. Find the recoil velocity of the gun.
V = − mv / M = − (0.02 × 400) / 5 = −1.6 m s⁻¹
The gun recoils at 1.6 m s⁻¹, opposite to the bullet.
Example 4.4 — Perfectly inelastic collision
A 2 kg body moving at 6 m s⁻¹ collides with a 4 kg body at rest and they stick together. Find the common velocity and the loss of kinetic energy.
v = (2 × 6 + 0) / (2 + 4) = 2 m s⁻¹
Ki = ½ × 2 × 6² = 36 J ; Kf = ½ × 6 × 2² = 12 J
Loss = 24 J
4.3 Application of Newton's Laws
Method for solving problems:
Draw a clear free-body diagram for each body, showing only the forces acting on that body.
Choose a positive direction for each body along its acceleration.
Apply ΣF = ma to each body.
Use the constraint: a rope that stays taut gives the same speed and acceleration to the bodies it connects, and an ideal (light, frictionless) pulley leaves the tension unchanged.
4.3(a) Atwood's Machine (Two Masses over a Pulley)
Figure 4.4: Atwood's machine with m₁ > m₂
Acceleration and tension
m₁: m₁g − T = m₁a
m₂: T − m₂g = m₂a
Adding: (m₁ − m₂) g = (m₁ + m₂) a
a = (m₁ − m₂) g / (m₁ + m₂) T = 2 m₁ m₂ g / (m₁ + m₂)
4.3(b) Block on a Smooth Table Connected to a Hanging Mass
Figure 4.5: Mass m₁ on a smooth horizontal table pulled by a hanging mass m₂ through a light string over a pulley
Acceleration and tension
m₁ (horizontal): T = m₁a
m₂ (vertical): m₂g − T = m₂a
a = m₂ g / (m₁ + m₂) T = m₁ m₂ g / (m₁ + m₂)
4.3(c) Body on a Smooth Inclined Plane
Figure 4.6: Free-body diagram of a block on a smooth plane inclined at θ
Perpendicular to the plane: N = mg cos θ
Along the plane (down the slope): mg sin θ = ma
a = g sin θ
The acceleration does not depend on the mass. For θ = 90° the plane becomes vertical and a = g (free fall).
4.3(d) Apparent Weight in a Lift
Figure 4.7: Forces on a body inside a lift. The reading of a weighing scale is the normal reaction N
Motion of the lift
Equation
Apparent weight N
At rest or at constant velocity
N − mg = 0
mg
Accelerating upward (a)
N − mg = ma
m(g + a) (greater)
Accelerating downward (a)
mg − N = ma
m(g − a) (smaller)
Free fall (a = g), e.g. cable breaks
mg − N = mg
0 (weightlessness)
Example 4.5 — Atwood's machine
Masses of 5 kg and 3 kg hang over a light pulley (g = 9.8 m s⁻²). Find a and T.
a = (5 − 3)(9.8)/(5 + 3) = 2.45 m s⁻²
T = 2(5)(3)(9.8)/8 = 36.75 N
Example 4.6 — Block and hanging mass
A 4 kg block on a smooth table is joined by a string over a pulley to a 1 kg mass hanging at the edge. Find a and T.
a = (1)(9.8)/(4 + 1) = 1.96 m s⁻²
T = (4)(1)(9.8)/5 = 7.84 N
Example 4.7 — Lift
A 60 kg man stands on a weighing scale in a lift accelerating at 2 m s⁻². What does the scale read when the lift (a) goes up, (b) goes down? (g = 9.8 m s⁻²)
(a) N = m(g + a) = 60 × 11.8 = 708 N
(b) N = m(g − a) = 60 × 7.8 = 468 N
4.4 Moment, Torque and Equilibrium
4.4(a) Moment of a Force (Torque)
The moment of a force (torque) about a point is the turning effect of the force about that point. Its magnitude is the force times the perpendicular distance from the point to the line of action of the force:
Figure 4.8: Torque about the pivot O. The perpendicular distance d = r sin θ is measured to the line of action of F
τ = F × d = rF sin θ
In vector form: τ = r × F
SI unit: N m; dimensions [ML²T⁻²]. By convention, anticlockwise torque is positive and clockwise torque is negative. The torque is zero if the line of action passes through the pivot (d = 0) or if F ∥ r (θ = 0° or 180°). The torque is maximum when F ⊥ r. This is why a door is easier to open when pushed far from the hinge, and at right angles to the door.
4.4(b) Couple
Two equal and opposite parallel forces whose lines of action do not coincide form a couple. It produces rotation without translation, because the net force is zero.
Figure 4.9: A couple: equal and opposite forces F separated by perpendicular distance d
Moment of a couple = F × d
The moment of a couple is the same about any point. Examples: turning a steering wheel with both hands, opening a tap, turning a key.
4.4(c) Equilibrium of a Rigid Body
A rigid body is in equilibrium when it has no linear acceleration and no angular acceleration:
Conditions of equilibrium:
Translational: the vector sum of all external forces is zero, ΣF = 0.
Rotational: the sum of the torques about any point is zero, Στ = 0. This is the principle of moments: the total clockwise moment equals the total anticlockwise moment about any point.
Figure 4.10: A uniform beam of weight W resting on two supports A and B and carrying a load W₁
Reactions at the supports
For equilibrium of the beam in Fig 4.10 (weight W at the middle, load W₁ at distance x₁ from A):
Translational: RA + RB = W₁ + W
Torques about A: RB L = W₁ x₁ + W x₂ (x₂ = L/2 for a uniform beam)
RB = (W₁ x₁ + W x₂) / L , RA = W₁ + W − RB
Taking moments about the point where an unknown force acts removes that force from the equation. That is why A was chosen as the pivot.
Centre of gravity and stability:
The centre of gravity is the point where the whole weight of a body can be taken to act. For a uniform regular body it is the geometric centre.
Stable equilibrium: the body returns to its position after a small disturbance (centre of gravity rises).
Unstable equilibrium: the body moves further away (centre of gravity falls).
Neutral equilibrium: the body stays in its new position (centre of gravity stays at the same height), e.g. a ball on a level table.
A body with a low centre of gravity and a wide base is more stable.
Example 4.8 — Torque
A force of 20 N acts perpendicular to a spanner at 0.25 m from the nut. Find the torque.
τ = F d = 20 × 0.25 = 5 N m
Example 4.9 — Couple
Two forces of 10 N each, equal and opposite, act on a wheel at points 0.3 m apart. Find the moment of the couple.
Moment = F × d = 10 × 0.3 = 3 N m
Example 4.10 — Beam on two supports
A uniform beam of length 4 m and weight 100 N rests on supports at its two ends. A load of 200 N hangs 1 m from end A. Find the reactions at A and B.
Moments about A: RB × 4 = 200 × 1 + 100 × 2 = 400 ⇒ RB = 100 N
RA + RB = 300 ⇒ RA = 200 N
4.5 Solid Friction: Laws of Solid Friction and Their Verification
Friction is the force that opposes the relative motion, or the tendency of relative motion, between two surfaces in contact. It acts along the surfaces and arises from the roughness (and molecular adhesion) of the surfaces.
4.5(a) Types of Friction
Static friction (fs): acts when there is no relative motion. It adjusts itself to equal the applied force, up to a maximum value.
Limiting friction (fs,max): the maximum static friction, just when the body is about to slide.
Kinetic (sliding) friction (fk): acts when one surface slides over the other. It is slightly less than limiting friction.
Rolling friction: acts when a body rolls over a surface; it is much smaller than sliding friction.
Figure 4.11: Friction against applied force. Static friction rises equal to F up to the limiting value; once the body slides, friction drops to a smaller, nearly constant kinetic value
4.5(b) Laws of Solid Friction
Friction acts tangentially to the surfaces in contact and opposes the (impending) relative motion.
The limiting friction is independent of the apparent area of contact, as long as the normal reaction stays the same.
The limiting friction is directly proportional to the normal reaction N between the surfaces.
Kinetic friction is less than limiting friction, and is nearly independent of the speed of sliding (for moderate speeds).
Friction depends on the nature (material and roughness) of the two surfaces.
4.5(c) Coefficient of Friction
From law 3, the ratio of limiting friction to the normal reaction is a constant for a given pair of surfaces. It has no unit:
Coefficient of static friction: μs = fs,max / N
Coefficient of kinetic friction: μk = fk / N (μk < μs)
f ≤ μ N for static friction ; f = μk N when sliding
4.5(d) Angle of Friction and Angle of Repose
The angle of friction λ is the angle between the normal reaction and the resultant of the normal reaction and limiting friction: tan λ = f/N = μ. The angle of repose α is the maximum angle of an inclined plane at which a body placed on it just remains at rest.
Figure 4.12: A block on a rough inclined plane at the angle of repose: friction f acts up the slope
Angle of repose equals angle of friction
At the limiting angle α the block is about to slide, so friction takes its maximum value:
Along the plane: f = mg sin α
Perpendicular to the plane: N = mg cos α
μ = f / N = mg sin α / mg cos α
μ = tan α (so α = λ)
4.5(e) Verification of the Laws of Friction
Method 1: Inclined plane (angle of repose)
Place a rectangular wooden block on a plane whose inclination can be changed. Raise the plane slowly until the block just begins to slide. Measure the angle α (from the height and length of the plane); then μ = tan α.
Law 2: Repeat with the block resting on a smaller face (different area). The angle α, hence μ, is almost unchanged.
Law 3: Add weights on the block (N increases). The block still slides at the same α, so μ = f/N stays constant, which shows f ∝ N.
Law 5: Change the surface (rubber, glass, wood) and α changes.
Method 2: Horizontal surface with a pulley
Figure 4.13: Block of mass M on a rough horizontal surface pulled by a pan loaded with weights through a string over a pulley
Put weights on the pan until the block just begins to move (limiting friction), or gives a small push and moves with uniform velocity (kinetic friction).
Then the tension T = mg balances friction: f = mg, and N = Mg.
μ = f / N = m / M.
Law 3: Place extra masses on the block (M increases) and find the new m. A graph of m against M is a straight line through the origin, so f ∝ N.
Law 2: Turn the block on a different face; m remains almost the same.
4.5(f) Advantages, Disadvantages and Reduction of Friction
Useful: walking, braking, holding objects, nails and screws gripping, belts on pulleys.
Harmful: wear and tear, heat loss and waste of energy in machines.
Reduced by: lubricants (oil, grease), ball and roller bearings (rolling in place of sliding), polishing surfaces, streamlining.
Example 4.11 — Static and kinetic friction
A 5 kg block rests on a horizontal floor with μs = 0.4 and μk = 0.3 (g = 9.8 m s⁻²). Find the friction if the block is pushed with (a) 15 N, (b) 25 N, and the acceleration in case (b).
N = mg = 49 N ; fs,max = 0.4 × 49 = 19.6 N
(a) 15 N < 19.6 N: the block stays at rest and f = 15 N (static)
(b) 25 N > 19.6 N: the block slides; fk = 0.3 × 49 = 14.7 N
a = (25 − 14.7) / 5 = 2.06 m s⁻²
Example 4.12 — Angle of repose
A block just begins to slide when a plane is tilted to 25°. Find the coefficient of static friction.
μ = tan 25° = 0.466
Example 4.13 — Horizontal method
A 2 kg block moves with uniform velocity when the pan with weights has mass 0.6 kg. Find μk.
μk = m / M = 0.6 / 2 = 0.3
Key Formulas and Summary
Momentum and Impulse:
p = mv , F = dp/dt
J = F Δt = Δp = mv − mu
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ (no external force)
Atwood: a = (m₁ − m₂)g/(m₁ + m₂) , T = 2m₁m₂g/(m₁ + m₂)
Table + hanging mass: a = m₂g/(m₁ + m₂) , T = m₁m₂g/(m₁ + m₂)
Smooth incline: a = g sin θ
Lift: N = m(g ± a)
Torque, Equilibrium and Friction:
τ = rF sin θ = F d , couple = F d
ΣF = 0 , Στ = 0
f = μN , μ = tan α (angle of repose)
Horizontal method: μ = m/M
Practice Numerical Problems
Take g = 9.8 m s⁻² unless stated otherwise. Attempt each problem before reading the answer.
1. Find the momentum of a 1000 kg car moving at 72 km h⁻¹.
Answer: v = 20 m s⁻¹; p = 2 × 10⁴ kg m s⁻¹.
2. A force of 50 N acts for 0.2 s on a 0.5 kg ball at rest. Find the impulse and the final speed.
Answer: J = 50 × 0.2 = 10 N s; v = J/m = 20 m s⁻¹.
3. A 0.2 kg ball strikes a wall at 10 m s⁻¹ and rebounds at 10 m s⁻¹. The contact time is 0.02 s. Find the average force on the ball.
Answer: Δp = 0.2 × (10 + 10) = 4 N s; F = 4/0.02 = 200 N.
4. A rifle of mass 3 kg fires a 10 g bullet at 300 m s⁻¹. Find the recoil velocity of the rifle.
Answer: V = 0.01 × 300/3 = 1 m s⁻¹ (opposite to the bullet).
5. A 3 kg body moving at 4 m s⁻¹ collides with a 1 kg body at rest and they stick together. Find the common velocity and the loss of kinetic energy.
Answer: v = 12/4 = 3 m s⁻¹; Ki = 24 J, Kf = ½(4)(9) = 18 J; loss = 6 J.
6. A 2 kg ball moving at 5 m s⁻¹ collides elastically head-on with a stationary 2 kg ball. Find the velocities after the collision.
10. A 5 kg block slides down a smooth plane inclined at 37° (sin 37° = 0.6). Find its acceleration and the normal reaction.
Answer: a = g sin θ = 5.88 m s⁻²; N = mg cos θ = 5 × 9.8 × 0.8 = 39.2 N.
11. A force of 40 N acts at 30° to a 0.5 m rod, at its end, about the other end. Find the torque.
Answer: τ = rF sin θ = 0.5 × 40 × 0.5 = 10 N m.
12. A uniform plank 6 m long weighing 120 N rests on supports at its ends. A 180 N child stands 2 m from the left end. Find the reactions.
Answer: About the left end: RB × 6 = 180 × 2 + 120 × 3 = 720 ⇒ RB = 120 N; RA = 300 − 120 = 180 N.
13. A 20 kg box rests on a floor with μs = 0.3. What force just starts it moving?
Answer: fs,max = 0.3 × 20 × 9.8 = 58.8 N.
14. A block just slides on a plane when it is tilted at 30°. Find μ and the friction force on a 2 kg block at that angle.
Answer: μ = tan 30° ≈ 0.577; f = mg sin 30° = 2 × 9.8 × 0.5 = 9.8 N.
15. In the horizontal friction experiment a 4 kg block moves uniformly when the pan and weights total 1.2 kg. Find μk and the friction force.
Answer: μk = 1.2/4 = 0.3; f = mg = 1.2 × 9.8 = 11.76 N.
Conceptual and Long Answer Questions
1. Why does a cricketer move the hands backward while catching a fast ball?
The same change of momentum takes place over a longer time, so the average force on the hands (F = Δp/Δt) is smaller.
2. Is momentum conserved in an inelastic collision? Is kinetic energy?
Momentum is conserved (no external force); kinetic energy is not, as some is converted to heat, sound and deformation.
3. Why does a gun recoil when it fires a bullet?
Momentum is conserved. The bullet gains forward momentum, so the gun gains an equal and opposite momentum. Its recoil speed is small because its mass is large.
4. A person standing on a weighing scale in a lift feels lighter in a downward-accelerating lift. Explain.
N = m(g − a) < mg, so the scale reading (apparent weight) is less than the true weight. The true weight mg does not change.
5. Why is it easier to open a door by pushing at the handle than near the hinge?
The torque τ = F d increases with the perpendicular distance d from the hinge, so less force gives the same torque.
6. A couple has zero resultant force. Why does it still turn a body?
The two forces have different lines of action, so their torques add and the net torque F × d is not zero; there is rotation without translation.
7. Why is kinetic friction less than limiting static friction?
Once sliding begins, the surface irregularities have less time to interlock and bond, so the resisting force decreases (Fig 4.11).
8. State the laws of limiting friction. How would you verify that friction is independent of the area of contact? (Section 4.5)
9. Prove that the angle of repose is equal to the angle of friction. (Section 4.5)
10. Derive the law of conservation of linear momentum from Newton's laws, and obtain the velocities after an elastic collision. (Section 4.2)
Multiple Choice Questions
1. The SI unit of impulse is:
(a) N
(b) N s ✓
(c) N m
(d) J
2. In every collision, if no external force acts, which quantity is conserved?
(a) Kinetic energy
(b) Total linear momentum ✓
(c) Velocity of each body
(d) Speed of each body
3. A 2 kg ball at 3 m s⁻¹ hits an identical stationary ball elastically head-on. After the collision, the first ball:
(a) Moves at 3 m s⁻¹
(b) Moves at 1.5 m s⁻¹
(c) Comes to rest ✓
(d) Rebounds at 3 m s⁻¹
4. The apparent weight of a person in a lift moving upward with acceleration a is:
(a) mg
(b) m(g − a)
(c) m(g + a) ✓
(d) zero
5. The torque is zero when the angle between r and F is:
(a) 90°
(b) 45°
(c) 60°
(d) 0° ✓
6. A rigid body is in equilibrium when:
(a) ΣF = 0 only
(b) Στ = 0 only
(c) ΣF = 0 and Στ = 0 ✓
(d) Its velocity is zero
7. Limiting friction between two surfaces is directly proportional to:
(a) The area of contact
(b) The normal reaction ✓
(c) The speed of sliding
(d) The volume of the body
8. A block just slides on a plane inclined at 45°. The coefficient of static friction is:
Physics Decoded
Grade XI › Dynamics
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