These eight original questions mix the four lessons. Write your answer first, then open the explanation. Use g = 10 m s⁻² throughout.
If you need a lesson, start from the model-checking hub.
Questions
Question 1. A 50 kg skater and a 30 kg friend stand at rest on smooth ice. The friend moves off at 2.0 m s⁻¹. State the boundary and find the skater’s velocity.
Answer
Boundary: both people, with negligible friction, so momentum is conserved. 50v + 30(2.0) = 0 gives v = −1.2 m s⁻¹, which is 1.2 m s⁻¹ opposite to the friend.
Question 2. A 2.0 kg block slides down a rough slope from a height of 1.5 m and reaches the bottom at 3.0 m s⁻¹. Is mechanical energy conserved? Quantify your answer.
Answer
Potential energy lost = 2.0 × 10 × 1.5 = 30 J. Kinetic energy gained = ½ × 2.0 × 9.0 = 9.0 J. The 21 J difference shows that it is not conserved; friction transfers 21 J to thermal energy.
Question 3. A cyclist rides round a circular path at a constant 6.0 m s⁻¹. Is the cyclist accelerating? Explain in two sentences.
Answer
Yes. The speed is constant, but the direction of the velocity changes continuously, so the velocity changes and there is an acceleration towards the centre of the circle.
Question 4. A runner completes a 400 m lap in 80 s. State the average speed and the average velocity.
Answer
Average speed = 400 ÷ 80 = 5.0 m s⁻¹. Displacement is zero, so the average velocity is 0 m s⁻¹.
Question 5. A student writes s = ut + ½at for a motion problem. Show with units that the formula is wrong.
Answer
The term ut has units m. The term ½at has (m s⁻²)(s) = m s⁻¹. A length and a velocity cannot be added, so the formula is wrong. The correct term is ½at².
Question 6. To find the speed of a 0.50 kg ball with 25 J of kinetic energy, a student writes v = √(2E × m). Identify the error with units, and find the correct speed.
Answer
The units of 2Em are kg² m² s⁻², whose square root is kg m s⁻¹, which is a momentum, not a speed. The correct form is v = √(2E ÷ m) = √(50 ÷ 0.50) = √100 = 10 m s⁻¹.
Question 7. A train moving at 30 m s⁻¹ brakes at 3.0 m s⁻². At what time is it 120 m from where it began to brake? Solve 120 = 30t − 1.5t² and say which root is valid.
Answer
Rearrange: 1.5t² − 30t + 120 = 0, so t² − 20t + 80 = 0. Then t = [20 ± √(400 − 320)] ÷ 2 = [20 ± 8.94] ÷ 2, so t = 5.53 s or t = 14.47 s.
The train stops at 30 ÷ 3.0 = 10 s, so t = 5.53 s is valid and t = 14.47 s is rejected. Check: 30 × 5.53 − 1.5 × 30.6 = 165.9 − 45.9 = 120 m.
Question 8. A ball is thrown upward at 20 m s⁻¹ from the ground. When is it 15 m above the ground? Decide whether both roots are valid.
Answer
15 = 20t − 5t², so t² − 4t + 3 = 0 and t = 1 s or t = 3 s. The ball lands at t = 4 s, so both roots are valid: at 1 s it is rising, and at 3 s it is falling.
If you got these wrong
Match each question to a lesson: 1 and 2 to the boundary lesson, 3 and 4 to constant speed versus constant velocity, 5 and 6 to the unit-check lesson, and 7 and 8 to the rejected-root lesson.
Record the check you missed with the mistake log and paper-error review tool. If you want a teacher to work through new questions with you, see online one-to-one Physics tuition.