Edexcel IAL Physics: Further Mechanics — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Further Mechanics revision notes

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Questions

1. Explain how momentum is conserved in a two-dimensional collision. [2]

2. State the relationship between relative speed of approach and separation for a perfectly elastic collision. [1]

3. A 0.15 kg ball moving east at 4.0 m s⁻¹ collides with a stationary 0.25 kg ball. After the collision the first ball moves at 2.0 m s⁻¹ at 30° north of east.

(a) Calculate the east and north components of the first ball’s final momentum. [3] (b) Use conservation of momentum to find the east and north components of the second ball’s momentum. [3] (c) Calculate the speed of the second ball. [2]

4. A conical pendulum: a 0.20 kg bob on a 0.90 m string moves in a horizontal circle with the string at 35° to the vertical. (g = 9.81 m s⁻²)

(a) Calculate the radius of the circle. [2] (b) Calculate the tension in the string. [3] (c) Calculate the speed of the bob. [3]

5. A 0.40 kg ball on a 0.65 m string is swung in a vertical circle.

(a) Calculate the minimum speed at the top for the ball to maintain a circular path. [3] (b) At that speed, calculate the tension at the bottom of the circle, assuming the speed there is 5.2 m s⁻¹. [3]

6. Explain why the centripetal force does no work, and what this means for the speed of an object in uniform circular motion. [3]

7. State the equation relating impulse to change in momentum, and name the practical (CORE PRACTICAL 9) that investigates this relationship. [2]

8. A 0.15 kg ball moving at 8.0 m s⁻¹ collides head-on with a stationary 0.25 kg ball. After the collision the first ball rebounds at 2.0 m s⁻¹.

(a) Find the velocity of the second ball after the collision. [3] (b) By calculating the total kinetic energy before and after, determine whether the collision is elastic. [3]

9. State the equation Ek = p²/2m, and explain when it is more useful than Ek = ½mv². [2]

10. State the equation linking angular velocity to the period of rotation, and name the practical (CORE PRACTICAL 10) used to analyse collisions between small spheres. [2]

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Answers

1. Momentum is a vector, so it is conserved independently in each of two perpendicular directions [1]; resolve into components and apply conservation to each separately [1].

2. They are equal [1].

3. (a) p = 0.15 × 2.0 = 0.30 kg m s⁻¹ [1] East: 0.30 cos 30° = 0.260 [1]; North: 0.30 sin 30° = 0.150 kg m s⁻¹ [1]. (b) Initial momentum: east 0.15 × 4.0 = 0.60, north 0 [1] Second ball east: 0.60 − 0.260 = 0.340 [1]; north: 0 − 0.150 = −0.150 kg m s⁻¹ [1]. (c) |p| = √(0.340² + 0.150²) = √(0.1156 + 0.0225) = 0.372 [1] v = 0.372 ÷ 0.25 = 1.49 m s⁻¹ [1].

4. (a) r = L sin θ = 0.90 × sin 35° [1] = 0.516 m [1]. (b) Vertically: T cos θ = mg [1] T = (0.20 × 9.81) ÷ cos 35° [1] = 2.40 N [1]. (c) Horizontally: T sin θ = mv² ÷ r [1] 2.40 × sin 35° = (0.20 × v²) ÷ 0.516 [1] 1.376 = 0.3876v², v² = 3.55, v = 1.88 m s⁻¹ [1].

5. (a) At minimum speed T = 0, so weight alone provides the centripetal force [1] mg = mv² ÷ r, v = √(gr) = √(9.81 × 0.65) [1] = 2.53 m s⁻¹ [1]. (b) At the bottom: T − mg = mv² ÷ r [1] T = (0.40 × 5.2² ÷ 0.65) + (0.40 × 9.81) [1] = 16.64 + 3.92 = 20.6 N [1].

6. The force acts perpendicular to the velocity at every instant [1], and work requires a component of force along the displacement [1]. Since no work is done, the kinetic energy and hence the speed remain constant [1].

7. Impulse = FΔt = Δp [1] — CORE PRACTICAL 9 investigates the relationship between the force exerted on an object and its resulting change of momentum [1].

8. (a) Taking the initial direction as positive: (0.15 × 8.0) + (0.25 × 0) = (0.15 × −2.0) + (0.25 × v) [1]. 1.2 = −0.30 + 0.25v [1]. v = 1.5 ÷ 0.25 = 6.0 m s⁻¹ [1]. (b) KE before = ½(0.15)(8.0)² = 4.8 J [1]. KE after = ½(0.15)(2.0)² + ½(0.25)(6.0)² = 0.3 + 4.5 = 4.8 J [1]. Kinetic energy is conserved, so the collision is elastic [1].

9. Ek = p²/2m [1] — useful when momentum, rather than velocity, is known directly, avoiding the need to find v first before calculating kinetic energy [1].

10. ω = 2π/T [1] — CORE PRACTICAL 10 uses ICT, such as video analysis software, to analyse collisions between small spheres, for example ball bearings on a table top [1].

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Where marks are usually lost

• Not resolving momentum into components in a two-dimensional collision.
• Forgetting that both tension and weight act towards the centre at the top of a vertical circle.
• Using the string length instead of the radius in a conical pendulum.
• Saying the centripetal force changes the speed.
• Forgetting that Ek = p²/2m is derived from Ek = ½mv² and p = mv, and is chosen specifically because momentum, not velocity, is given.
• Confusing CORE PRACTICAL 9 (force and change of momentum) with CORE PRACTICAL 10 (ICT analysis of sphere collisions).