These eight original questions follow the lessons in force and motion I. Write your answer first, then open the explanation. Use g = 10 m s⁻² throughout.
If you need a lesson, start from the force and motion I hub and choose the skill that matches your mistakes.
Questions
Question 1. A train moves at a constant 20 m s⁻¹ for 30 s, then slows uniformly to rest in 10 s. Find the total displacement.
Answer
Constant stage: 20 × 30 = 600 m. Braking stage: area of a triangle = ½ × 10 × 20 = 100 m. Total = 700 m. See interpreting displacement, velocity and acceleration.
Question 2. A runner accelerates from rest at 2.5 m s⁻² for 6.0 s. Find the final speed and the distance covered.
Answer
Given u = 0, a = 2.5, t = 6.0. Use v = u + at: v = 15 m s⁻¹.
Use s = ut + ½at²: s = ½ × 2.5 × 36 = 45 m. Check: s = ½(u + v)t = ½ × 15 × 6 = 45 m.
Question 3. A car at 25 m s⁻¹ brakes at 5.0 m s⁻². How far does it travel before it stops?
Answer
Take forward as positive: u = 25, v = 0, a = −5. Time is missing, so use v² = u² + 2as: 0 = 625 − 10s, so s = 62.5 m.
Question 4. A 0.50 kg trolley at 4.0 m s⁻¹ collides with a stationary 1.5 kg trolley and they stick together. Find their common speed.
Answer
Momentum before = 0.50 × 4.0 = 2.0 kg m s⁻¹. After: 2.0v = 2.0, so v = 1.0 m s⁻¹. See explaining momentum and impulse.
Question 5. A 60 kg person lands from a jump and stops in 0.25 s after hitting the ground at 4.0 m s⁻¹. Find the average force of the ground on the person, ignoring weight.
Answer
Impulse = change in momentum = 60 × 4.0 = 240 N s. Force = 240 ÷ 0.25 = 960 N. Bending the knees lengthens the time, which would lower the force.
Question 6. A 5.0 kg box is pushed along a floor by a 30 N force. Friction is 10 N. Find the acceleration.
Answer
Resultant force = 30 − 10 = 20 N. From F = ma, a = 20 ÷ 5.0 = 4.0 m s⁻² in the direction of the push. See applying Newton’s laws in context.
Question 7. A 2.0 kg object is lifted 3.0 m at constant speed in 4.0 s. Find the work done and the power.
Answer
Force = weight = 2.0 × 10 = 20 N. Work = 20 × 3.0 = 60 J. Power = 60 ÷ 4.0 = 15 W. See distinguishing work, energy and power.
Question 8. A trolley slows from 6.0 m s⁻¹ at 2.0 m s⁻². A student finds its position at 5.0 s as s = 6 × 5 − ½ × 2 × 25 = 5 m. Identify the flaw and give the correct position.
Answer
The trolley stops at t = 3.0 s, since 6 − 2t = 0. After that it does not reverse, so the formula no longer applies.
Distance at stopping = 6 × 3 − ½ × 2 × 9 = 9 m. The position at 5.0 s is 9 m. See selecting and checking a physical model.
If you got these wrong
Match each question to a lesson: 1 to the graph lesson, 2 and 3 to the equations lesson, 4 and 5 to momentum and impulse, 6 to Newton’s laws and drawing force diagrams, 7 to work and power, and 8 to the model-checking section.
Record the error type with the mistake log and paper-error review tool. If you want a teacher to review your working with you, see online one-to-one Physics tuition.