S4 Physics Practice Paper: Force, Motion and Energy (100 marks)

S4 Physics · Mechanics: force, motion and energy

S4 Physics Practice Paper: Force, Motion and Energy (100 marks): 6 questions, 26 parts, 100 marks, 75 minutes.

Section A — Uniform acceleration and velocity-time graphs

Show your working and state the unit in each answer.

#	Question	Answer and marking notes
1	A car starts from rest and accelerates at 2.5 m/s² for 8 s. Find its final velocity.	v = u + at = 0 + 2.5 × 8 = 20 m/s
2	What is the displacement of the car during these 8 s?	s = ½ × 2.5 × 8² = 80 m
3	The car then travels at a constant 20 m/s for 12 s. Find this displacement.	s = v × t = 20 × 12 = 240 m
4	The car finally decelerates uniformly at 4 m/s² to rest. Find the time and distance.	t = 20 ÷ 4 = 5 s; s = ½ × 20 × 5 = 50 m
5	Find the average velocity of the whole journey and state what the area under the v-t graph represents.	Total displacement = 80 + 240 + 50 = 370 m; average velocity = 370 ÷ 25 = 14.8 m/s; the area under the graph equals the displacement (370 m)

Take g = 10 m/s². Show your working.

#	Question	Answer and marking notes
6	A 4 kg object on a horizontal surface is pulled by an 18 N horizontal force against a friction of 6 N. Find its acceleration.	Net force = 18 − 6 = 12 N; a = 12 ÷ 4 = 3 m/s²
7	Starting from rest, what is its velocity after 5 s?	v = at = 3 × 5 = 15 m/s
8	Find the coefficient of friction between the object and the table.	μ = f ÷ N = 6 ÷ (4 × 10) = 0.15
9	If the pull rises to 26 N with unchanged friction, find the new acceleration.	Net force = 26 − 6 = 20 N; a = 20 ÷ 4 = 5 m/s²

Take g = 10 m/s². State the unit in each answer.

#	Question	Answer and marking notes
10	A 20 kg box is pushed to the top of a slope of length 5 m rising 3 m. Find the gain in potential energy.	ΔPE = mgh = 20 × 10 × 3 = 600 J
11	If friction is 25 N, find the work done against friction.	W = f × d = 25 × 5 = 125 J
12	Find the total work done in pushing the box.	Total work = 600 + 125 = 725 J
13	If the motor delivers 300 W, how long does the process take (3 significant figures)?	t = W ÷ P = 725 ÷ 300 = 2.42 s
14	Find the efficiency of this process.	η = 600 ÷ 725 = 0.828 = 82.8%

Ignore air resistance and take g = 10 m/s².

#	Question	Answer and marking notes
15	A 0.5 kg ball falls freely from a height of 1.8 m. Find its speed on landing.	v = √(2 × 10 × 1.8) = 6 m/s
16	Find the kinetic energy of the ball just before landing.	KE = ½ × 0.5 × 6² = 9 J
17	The ball rebounds to a height of 1.25 m. Find its initial speed after the bounce.	v = √(2 × 10 × 1.25) = 5 m/s
18	Find the energy lost during the impact, expressed as a percentage (1 decimal place).	Loss = 0.5 × 10 × (1.8 − 1.25) = 2.75 J; 2.75 ÷ 9 = 30.6%
19	If 80% of the energy is retained at each bounce, find the height of the second rebound.	1.25 × 0.8 = 1.0 m

The ruler stays horizontal. Show your working and the unit.

#	Question	Answer and marking notes
20	A uniform 1.2 m ruler weighs 4 N and is pivoted at its centre. Find the moment of the ruler about the pivot.	Its centre of gravity is directly above the pivot, so the lever arm is zero; moment = 0 N·m
21	A 6 N weight hangs from the left end. Find its moment about the pivot.	Moment = 0.6 × 6 = 3.6 N·m (anticlockwise)
22	What weight must hang 0.4 m to the right of the pivot for balance?	W = 3.6 ÷ 0.4 = 9 N
23	Find the total downward force on the pivot.	4 + 6 + 9 = 19 N

State the unit; give angles to 1 decimal place.

#	Question	Answer and marking notes
24	Two forces of 12 N and 16 N act at right angles. Find the magnitude of the resultant.	R = √(12² + 16²) = √400 = 20 N
25	Find the angle between the resultant and the 12 N force.	tan θ = 16 ÷ 12; θ = 53.1°
26	If the two forces act in opposite directions, what is the resultant?	16 − 12 = 4 N, along the direction of the 16 N force

A TutorZone practice paper; original questions and passages, not publisher material.