Forces -- Diagnostic Tests

Forces — Diagnostic Tests

Unit Tests

UT-1: Newton’s Laws and Resultant Forces

Question: (a) State Newton’s three laws of motion. (b) A 5 kg object is acted on by two forces: 30 N to the right and 20 N to the left. Calculate the acceleration. (c) A person of mass 70 kg stands in a lift. Calculate the normal force when: (i) the lift is stationary, (ii) accelerating upwards at $2 \text{ m/s^2$ (iii) accelerating downwards at $3 \text{ m/s^2$ . (d) Explain why the person feels heavier when the lift accelerates upwards.

Solution:

(a) First Law: An object remains at rest or moves at constant velocity unless acted on by a resultant (net) force. Second Law: The acceleration of an object is directly proportional to the resultant force and inversely proportional to its mass: $F = ma$ . Third Law: When two objects interact, they exert equal and opposite forces on each other.

(b) Resultant force $= 30 - 20 = 10$ N to the right. $a = F/m = 10/5 = 2$ m/s $^2$ to the right.

(c) Weight $= 70 \times 9.8 = 686$ N. (i) Stationary: $N = 686$ N. (ii) Accelerating up: $N - 686 = 70 \times 2 = 140$ . $N = 826$ N. (iii) Accelerating down: $686 - N = 70 \times 3 = 210$ . $N = 476$ N.

(d) When the lift accelerates upwards, the floor must push up on the person with a force greater than their weight to accelerate them upwards (Newton’s 2nd law: $N - W = ma$ ). The increased normal force is what the person perceives as feeling heavier. Their apparent weight (what a scales under their feet would read) increases.

UT-2: Hooke’s Law and Elasticity

Question: (a) State Hooke’s law and explain its limitations. (b) A spring has an original length of 12 cm. When a 4 N weight is hung from it, the length becomes 16 cm. Calculate the spring constant. (c) Calculate the extension when a 7 N weight is hung from the same spring. (d) Explain the difference between elastic and inelastic deformation with examples.

Solution:

(a) Hooke’s law states that the extension of a spring is directly proportional to the force applied, provided the limit of proportionality is not exceeded: $F = kx$ Where $k$ is the spring constant and $x$ is the extension.

Limitation: Hooke’s law only applies up to the limit of proportionality (elastic limit). Beyond this point, the extension is no longer proportional to the force, and the spring may be permanently deformed.

(b) Extension $x = 16 - 12 = 4$ cm $= 0.04$ m. $k = F/x = 4/0.04 = 100$ N/m.

(d) Elastic deformation: The object returns to its original shape when the force is removed. The energy stored is fully recovered. Example: a rubber band stretched within its elastic limit, a spring compressed slightly.

Inelastic (plastic) deformation: The object does not return to its original shape when the force is removed. It is permanently deformed, and some energy is used to change the object’s shape (not recoverable). Example: a paperclip bent out of shape, modelling clay, a spring stretched beyond its elastic limit.

UT-3: Momentum and Pressure

Question: (a) Define momentum and state its unit. (b) A 1500 kg car travelling at 20 m/s collides with a stationary 1000 kg car and they stick together. Calculate the velocity of the combined wreck immediately after the collision. (c) Calculate the pressure exerted by a 60 kg person standing on both feet, each foot having an area of $0.02 \text{ m^2$ . (d) Explain why the pressure exerted by a person wearing stiletto heels is much greater than when wearing flat shoes.

Solution:

(a) Momentum $=$ mass $\times$ velocity. It is a vector quantity. Unit: kg m/s.

(b) Before: $p = 1500 \times 20 + 1000 \times 0 = 30,000$ kg m/s. After: $p = (1500 + 1000) \times v = 2500v$ . Conservation of momentum: $2500v = 30,000$ . $v = 12$ m/s.

(c) Total area $= 2 \times 0.02 = 0.04$ m $^2$ . Weight $= 60 \times 9.8 = 588$ N. Pressure $= F/A = 588/0.04 = 14,700$ Pa $= 14.7$ kPa.

(d) Pressure $= F/A$ . Stiletto heels have a very small contact area (perhaps $1 \text{ cm^2 = 0.0001 \text{ m^2$ ), so the same weight (force) is distributed over a much smaller area, producing a much higher pressure. $P = 588/0.0001 = 5,880,000$ Pa $\approx 5.9$ MPa, compared to 14.7 kPa with flat shoes. This is why stiletto heels can damage wooden floors.

Integration Tests

IT-1: Forces in Circular Motion (with Energy)

Question: A car of mass 1200 kg travels around a roundabout of radius 25 m at a constant speed of 12 m/s. (a) Calculate the centripetal acceleration. (b) Calculate the centripetal force and explain what provides this force. (c) The maximum frictional force between the tyres and road is 8000 N. Calculate the maximum safe speed. (d) Explain what would happen if the car exceeded this speed.

Solution:

(a) $a = v^2/r = 144/25 = 5.76$ m/s $^2$ .

(b) $F = mv^2/r = 1200 \times 144/25 = 1200 \times 5.76 = 6912$ N. This force is provided by friction between the tyres and the road surface. Friction acts horizontally towards the centre of the roundabout, providing the centripetal force needed to keep the car moving in a circle.

(c) $F_{\text{max} = mv_{\text{max}^2/r$ . $8000 = 1200 \times v_{\text{max}^2/25$ . $v_{\text{max}^2 = 8000 \times 25/1200 = 166.67$ . $v_{\text{max} = 12.91$ m/s.

(d) If the car exceeds the maximum safe speed, the required centripetal force exceeds the maximum available friction. The car cannot maintain its circular path and will skid outwards (slide off the roundabout). This is because the friction is insufficient to provide the centripetal acceleration needed for the circular path at that speed.

IT-2: Stopping Distance and Braking (with Energy)

Question: A car of mass 1000 kg is travelling at 20 m/s. The driver’s reaction time is 0.7 seconds. (a) Calculate the thinking distance. (b) The braking force is 5000 N. Calculate the braking distance using $F = ma$ and kinematics. (c) Calculate the total stopping distance. (d) Explain how stopping distance changes if the road is wet (coefficient of friction halved) and if the car is loaded with an extra 300 kg.

Solution:

(a) Thinking distance $= \text{speed \times \text{reaction time = 20 \times 0.7 = 14$ m.

(b) $a = F/m = 5000/1000 = 5$ m/s $^2$ (deceleration). Using $v^2 = u^2 + 2as$ : $0 = 400 + 2(-5)s$ . $10s = 400$ . $s = 40$ m.

(d) Wet road (friction halved): Braking force halves to 2500 N. Deceleration $= 2.5$ m/s $^2$ . Braking distance $= 400/5 = 80$ m (doubles). Total $= 14 + 80 = 94$ m. Thinking distance is unchanged (reaction time and speed are the same).

Loaded car (1300 kg): Braking force unchanged at 5000 N. Deceleration $= 5000/1300 = 3.85$ m/s $^2$ . Braking distance $= 400/7.69 = 52$ m. Total $= 14 + 52 = 66$ m. The increased mass means the same braking force produces less deceleration, increasing stopping distance.

IT-3: Free Fall and Terminal Velocity (with Particles)

Question: A skydiver of mass 80 kg jumps from a plane. (a) Calculate the initial acceleration (ignore air resistance). (b) Explain how air resistance changes as the skydiver falls. (c) At terminal velocity, the skydiver falls at 55 m/s. Calculate the air resistance at this speed. (d) When the parachute opens, the terminal velocity drops to 5 m/s. Calculate the new air resistance and explain what happens to the skydiver’s motion immediately after the parachute opens.

Solution:

(a) Initial acceleration $= g = 9.8$ m/s $^2$ (only gravity acts; air resistance is zero at the instant of jumping).

(b) As the skydiver accelerates, their speed increases. Air resistance (drag) increases with speed ( $F_d \propto v^2$ ). As air resistance increases, the resultant downward force decreases ( $F_{\text{net} = W - F_d$ ), so acceleration decreases. The skydiver accelerates less and less until air resistance equals weight.

(c) At terminal velocity: air resistance $=$ weight $= 80 \times 9.8 = 784$ N. The resultant force is zero and acceleration is zero.

(d) New terminal velocity $= 5$ m/s. At this new terminal velocity, the drag force still equals weight $= 784$ N. However, the parachute provides much more drag at the same speed, so terminal velocity is much lower.

Immediately after opening: the parachute suddenly increases the drag force dramatically (well above 784 N). The resultant force is now upward ( $F_d \gt W$ ), causing the skydiver to decelerate rapidly. As speed decreases, drag decreases until it equals weight again at the new, lower terminal velocity of 5 m/s.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

Confusing terminology or concepts that appear similar but have distinct meanings.
Overlooking key assumptions or boundary conditions that limit applicability.

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