Topic 3: Motion and Forces
Course: Prep4Uni Physics 1
Chapter 1: Quantities and Measurement
Chapter 2: Forces and Moments
Chapter 3: Motion and Forces
Chapter 4: Energy and Fields
Chapter 5: Projectile Motion
🚁Overview
This topic develops your understanding of how and why objects move. It connects the description of motion (kinematics) with the forces that cause motion, using Newton’s Laws as a foundation for analyzing dynamics in real-world and theoretical situations.
📖Contents
Kinematics and uniformly accelerated motion
Linear momentum
Newton’s Laws of Motion
🎯Learning Outcomes
By the end of this section, students should be able to:
Define and use terms: displacement, speed, velocity, and acceleration
Use and interpret motion-time and velocity-time graphs
Derive and apply the kinematic equations for uniformly accelerated motion
Understand mass as a measure of inertia
Define and use momentum in practical situations
Apply Newton’s Laws to explain and predict motion
Solve problems using Newton’s Second Law: F = ma
Table of Contents
🔢1. Kinematics and Uniformly Accelerated Motion
Key Concepts:
Displacement (s): The change in position (vector)
Speed (v): Scalar rate of motion; v = distance/time
Velocity (v): Vector version of speed; includes direction
Acceleration (a): Rate of change of velocity; a = Δv/Δt
Equations of Motion (constant acceleration):
v = u + at
s = ut + ½at²
v² = u² + 2as
(Where u = initial velocity, v = final velocity, a = acceleration, s = displacement, t = time)
Example:
A car accelerates from 10 m/s to 30 m/s in 5 seconds. Find its acceleration.
a = (v – u)/t = (30 – 10)/5 = 4 m/s²
📐Learning-Check 1:
Q: A ball is thrown upward at 15 m/s. How long until it reaches its highest point?
A: At highest point, v = 0 → t = (v – u)/a = (0 – 15)/(-9.8) ≈ 1.53 s
📐Learning-Check 2:
Q: A train accelerates uniformly from rest to 25 m/s in 10 s. Find displacement.
A: s = ut + ½at² → a = 25/10 = 2.5; s = 0 + ½ × 2.5 × 100 = 125 m
🔢2. Interpreting Motion Graphs
Displacement–Time Graphs:
Slope = velocity
Displacement Time Graph Velocity–Time Graphs:
Slope = acceleration; Area under graph = displacement
Example:
A velocity–time graph shows a straight line rising from 0 to 20 m/s in 4 s.
Acceleration = slope = Δv / Δt = 20 / 4 = 5 m/s²
Displacement = area under triangle = ½ × base × height = ½ × 4 × 20 = 40 m
📐Learning-Check 1:
Q: On a velocity-time graph, what does a horizontal line represent?
A: Constant velocity (zero acceleration)
📐Learning-Check 2:
Q: What does the area under a velocity-time graph represent?
A: Displacement
🔢3. Newton’s Laws of Motion
First Law (Inertia): An object remains at rest or moves with constant velocity unless acted on by a net external force.
Second Law: F = ma – Net force causes acceleration proportional to mass.
Third Law: For every action, there is an equal and opposite reaction.
Example:
A 3 kg object accelerates at 2 m/s². What is the net force?
F = ma = 3 × 2 = 6 N
📐Learning-Check 1:
Q: A box is pushed with 10 N across a frictionless surface. It accelerates at 5 m/s². What is its mass?
A: m = F/a = 10 / 5 = 2 kg
📐Learning-Check 2:
Q: Which law explains why passengers lurch forward when a car stops suddenly?
A: Newton’s First Law (Inertia)
Illustration Video:
🔢4. Mass and Inertia
Mass measures an object’s resistance to changes in motion.
More mass → greater inertia → harder to accelerate.
Inertia explains why you lurch forward in a braking car.
📐Learning-Check 1:
Q: Two carts, one light and one heavy, are pushed with equal force. Which accelerates more?
A: The lighter cart (less inertia)
📐Learning-Check 2:
Q: What property resists changes in velocity?
A: Inertia (mass)
🔢5. Linear Momentum
Momentum (p): p = mv
Impulse (Δp): Change in momentum = force × time
In collisions, total momentum is conserved (if no external force).
Example:
A 2 kg trolley moving at 3 m/s hits a wall and stops in 0.1 s.
Impulse = Δp = 2 × 3 = 6 kg·m/s →
Average force = Δp / Δt = 6 / 0.1 = 60 N
📐Learning-Check 1:
Q: A 0.5 kg ball rolls at 8 m/s. What is its momentum?
A: p = mv = 0.5 × 8 = 4 kg·m/s
📐Learning-Check 2:
Q: A force of 20 N acts for 0.2 s. What is the impulse delivered?
A: Impulse = F × t = 20 × 0.2 = 4 N·s
⚙️Key Concepts Recap
| Concept | Formula | Notes |
|---|---|---|
| Displacement | s = ut + ½at² | Uniform acceleration |
| Final velocity | v = u + at | |
| Newton’s 2nd Law | F = ma | Force causes acceleration |
| Momentum | p = mv | Linear momentum |
| Impulse | FΔt = Δp | Force × time = change in momentum |
Proceed to: Topic 4: Energy and Fields
Go back to Prep4Uni Physics 1
📝EXERCISES
25 Learning‐Check Questions & Answers
(5 from each of the 5 subsections in “Motion and Forces”)
Section 1: Displacement, Speed, Velocity & Acceleration
Q: What is the difference between distance and displacement?
A: Distance is the total path length (scalar); displacement is the straight‐line change in position from start to finish (vector).Q: A runner goes 200 m east then 150 m west. What are her total distance and displacement?
A: Distance = 350 m; Displacement = 200 m – 150 m = 50 m east.Q: Define speed and instantaneous velocity.
A: Speed = distance / time (scalar). Instantaneous velocity = rate of change of displacement at a moment (vector).Q: A car covers 180 km in 3 h due north, then 60 km in 1 h east. Find its average velocity vector (magnitude & direction).
A: Displacement = (0 km, 180 km)+(60 km, 0) = (60, 180) km; total time = 4 h.
Speed = √(60²+180²)/4 = √(3600+32400)/4 = √36000/4 = 189.74/4 ≈47.4 km/h.
Direction = arctan(180/60)=71.6° north of east.Q: What is the physical meaning of a negative acceleration?
A: The velocity vector’s magnitude is decreasing (deceleration) or the velocity is reversing direction.
Section 2: Motion Graphs
Q: On a displacement–time graph, how do you tell if an object is accelerating?
A: The curve is non‐linear (slope changing). Constant slope = constant velocity; changing slope = acceleration.Q: On a velocity–time graph, what does the area under the curve represent?
A: Displacement over that time interval.Q: A v–t graph is a straight line from (t=0, v=0) to (t=5 s, v=20 m/s). What is the acceleration and distance traveled?
A: a = Δv/Δt = 20/5 = 4 m/s². Distance = area of triangle = ½·5·20 = 50 m.Q: What feature of a v–t graph indicates constant velocity?
A: A horizontal line (zero slope).Q: How do you find instantaneous acceleration on an s–t graph?
A: Compute the curvature; acceleration = second derivative d²s/dt². Graphically, a plot bending upwards indicates positive acceleration.
Section 3: Uniformly Accelerated Motion
Q: State the three kinematic (“suvat”) equations for constant acceleration.
A:v = u + a·t
s = u·t + ½·a·t²
v² = u² + 2·a·s
Q: Derive s = ½·a·t² for an object starting from rest.
A: With u=0, from s = u·t + ½·a·t² → s = 0 + ½·a·t².Q: How long does it take a stone dropped from rest to fall 20 m? (g=9.8 m/s²)
A: s = ½·gt² → t² = 2s/g = 40/9.8 ≈4.08 → t≈2.02 s.Q: A car accelerates uniformly from 10 m/s to 30 m/s over 50 m. Find its acceleration.
A: v²=u²+2as → 30²=10²+2a·50 →900=100+100a →a=8 m/s².Q: If acceleration is zero, which kinematic equations still hold?
A: v = u (constant velocity); s = u·t; v² = u².
Section 4: Linear Momentum
Q: Define linear momentum and its SI unit.
A: p = m·v; unit kg·m/s.Q: What is impulse, and what is its relation to momentum?
A: Impulse J = F·Δt; J = Δp (change in momentum).Q: A 0.5 kg ball thrown at 20 m/s has what momentum?
A: p = 0.5·20 = 10 kg·m/s.Q: A constant 5 N force acts for 3 s on a 1 kg mass initially at rest. What is its final speed?
A: Impulse = 5·3=15 N·s →Δp=15 →m·Δv=15 →v=15/1=15 m/s.Q: State conservation of momentum for a closed system.
A: Total momentum before = total momentum after, if no external forces.
Section 5: Newton’s Laws of Motion
Q: State Newton’s First Law.
A: A body stays at rest or moves at constant velocity unless acted upon by a net external force.Q: Give a real‐world example of Newton’s First Law.
A: A puck glides on ice at constant speed when friction is negligible.Q: State Newton’s Second Law.
A: ∑F = m·a (net force equals mass times acceleration).Q: What is the reaction force to the weight of a book on a table, according to Newton’s Third Law?
A: The book exerts a downward equal‐and‐opposite force on the table.Q: A 5 kg block is pushed with 20 N across a frictionless floor. What is its acceleration?
A: a = F/m = 20/5 = 4 m/s².
20 Problems with Detailed Solutions
(Covering all five sections of “Motion and Forces”)
Section 1: Displacement & Kinematics
Variable‐Speed Journey
A car’s velocity is v(t)=4t (m/s) for 0≤t≤5 s.
a) Find its displacement from t=0 to 5 s.
b) Find acceleration at t=2 s.
Solution:
a) s=∫₀⁵4t dt = [2t²]₀⁵ = 2·25 = 50 m.
b) a = dv/dt = 4 m/s² (constant).Change of Direction
A cyclist travels 100 m east in 20 s, then 60 m west in 15 s.
a) Total distance and displacement.
b) Average speed and average velocity magnitude.
Solution:
a) Distance=160 m; Displacement=100−60=40 m east.
b) Avg speed=160/35≈4.57 m/s; Avg velocity=40/35≈1.14 m/s east.Instantaneous vs. Average
A particle’s position is s(t)=t³–6t²+9t (m).
a) Find its average velocity between t=1 s and t=3 s.
b) Find its instantaneous velocity at t=2 s.
Solution:
a) s(3)=27−54+27=0; s(1)=1−6+9=4 →Δs=0−4=–4 m; Δt=2 s →v_avg=–2 m/s.
b) v(t)=3t²–12t+9 →v(2)=12−24+9=–3 m/s.Non‐Uniform Motion
A runner’s speed increases uniformly from 5 m/s to 9 m/s in 4 s, then remains constant.
a) Sketch v–t graph.
b) Compute distance run in first 4 s and next 6 s.
Solution:
a) Straight line from (0,5) to (4,9), then horizontal at v=9.
b) First 4 s: area trapezoid = ½(5+9)·4 = 28 m. Next 6 s: 9·6 = 54 m.Rapid Deceleration
A car moving at 25 m/s brakes with a constant deceleration of 5 m/s².
a) How long to stop?
b) How far does it travel during braking?
Solution:
a) v = u + at → 0 = 25 −5t → t=5 s.
b) s = u·t + ½at² = 25·5 + ½(−5)·25 = 125 −62.5 = 62.5 m.
Section 2: Motion Graphs
Piecewise v–t Analysis
Given v–t graph: 0→4 s at 2 m/s; 4→6 s linear to 6 m/s; 6→10 s constant.
Find displacement over 0–10 s.
Solution:
0–4: area =2·4=8; 4–6: area trapezoid=(2+6)/2·2=8; 6–10:6·4=24 →total=40 m.Reading s–t Curves
A displacement–time graph is a parabola s=2t².
a) What is the instantaneous velocity at t=3 s?
b) What is the acceleration?
Solution:
v=ds/dt=4t →v(3)=12 m/s; a=dv/dt=4 m/s².Acceleration–Time to Velocity
An a–t graph is horizontal at a=3 m/s² from t=0 to 5 s. If u=2 m/s, find v–t equation and distance traveled.
Solution:
v= u + at = 2 + 3t. Distance: ∫₀⁵(2+3t)dt = [2t +1.5t²]₀⁵ = 10 +37.5 = 47.5 m.Slope of s–t
A line on an s–t graph passes through (2, 6 m) and (5, 15 m). What is the speed?
A: slope = Δs/Δt = (15−6)/(5−2)=9/3=3 m/s.Curved vs. Straight
Explain why a curved s–t graph always implies non‐zero acceleration.
A: Because slope (velocity) varies with time.
Section 3: Uniformly Accelerated Motion
Uphill Skate
A skateboarder goes up a 5 m slope with u=6 m/s, coming to rest halfway. Find acceleration.
Solution:
v² = u² +2as → 0=36+2a·(2.5) → a=−36/5 = −7.2 m/s².Downward Throw
A stone is thrown downward at 5 m/s from 20 m. How long to hit ground?
Solution:
s=ut+½gt² →20=5t+4.9t² →4.9t²+5t−20=0 →t≈1.51 s.Upward Launch
A football is kicked at 15 m/s upward. Find:
a) Time to reach max height.
b) Max height.
Solution:
a) v= u−gt →0=15−9.8t →t=1.53 s.
b) s=ut−½gt²=15·1.53−4.9·(1.53)²≈11.48 m.Braking Distance
A car at 20 m/s decelerates at 4 m/s². How far to stop?
Solution:
v² = u² +2as →0=400+2·(−4)·s →s=400/8=50 m.Mid‐race Speed
A racer from rest accelerates at 3 m/s² for 8 s then coasts. Find speed at t=8 s.
A: v=3·8=24 m/s.
Section 4: Linear Momentum
Crash Test
A 1000 kg car moving at 20 m/s collides and stops in 0.1 s. Find the average force on the barrier.
Solution:
Δp = 1000·20 = 20000 kg·m/s; F_avg = Δp/Δt =20000/0.1=2×10⁵ N.Recoil of Gun
A 0.01 kg bullet leaves a 5 kg rifle at 400 m/s. Find the rifle’s recoil speed.
Solution:
0 = m_rifle·v_rifle + m_bullet·v_bullet →v_rifle = −(0.01·400)/5 = −0.8 m/s.Two‐Mass Collision
A 2 kg block moving at 3 m/s hits a stationary 1 kg block elastically. Find their speeds after.
Solution:
v₁’ = (m₁−m₂)/(m₁+m₂)·u₁ = (2−1)/3·3 = 1 m/s;
v₂’ = 2m₁/(m₁+m₂)·u₁ = 4/3·3=4 m/s.Inelastic Impact
A 0.2 kg ball at 5 m/s sticks to 0.8 kg block at rest. Find final velocity.
Solution:
(0.2·5 + 0.8·0)/1.0 =1 m/s.Rocket Propulsion (Simplified)
A rocket ejects 100 kg of gas at 2000 m/s. If initial mass is 1000 kg, what Δv?
Solution (Tsiolkovsky):
Δv = vₑ ln(m₀/m₁) =2000·ln(1000/900) ≈2000·0.1053≈210.6 m/s.
Section 5: Newton’s Laws
Atwood Machine
m₁=5 kg, m₂=3 kg over frictionless pulley. Find acceleration and tension.
Solution:
a = (m₁−m₂)g/(m₁+m₂) =2·9.8/8 =2.45 m/s².
T = m₁(g−a)=5·(9.8−2.45)=37.75 N.Block on Incline with Friction
m=4 kg, θ=30°, μₖ=0.2. Find acceleration down slope.
Solution:
F‖=mg sin30=19.6 N, f=μₖmg cos30=0.2·38.0=7.60 N →net=12.0 N →a=12/4=3 m/s².Centripetal Force
A 0.5 kg mass on 2 m string circles at 2 rev/s. Find required tension.
Solution:
ω=2·2π=4π rad/s; F= mω²r=0.5·(4π)²·2 ≈0.5·(16π²)·2=16π²≈158 N.Elevator Tension
Elevator m=600 kg accelerates upward at 1.5 m/s². Find cable tension.
Solution:
T−mg = ma →T = m(g+a)=600·(9.8+1.5)=6660 N.Forced Oscillation (Hooke + Newton)
Mass m=0.5 kg on spring k=100 N/m pulled down x=0.1 m and released. What is max acceleration?
Solution:
F_max = kx =100·0.1=10 N →a_max=F/m=10/0.5=20 m/s².
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