NCEA Level 3 Physics 91524 Mechanics free practice questions with answers

Question 1 (14 marks) [Translational motion and gravity]

(a) A 1.00 kg puck moving at 5.00 m s⁻¹ strikes a stationary 0.800 kg puck on a frictionless table. After the collision the first puck moves at 3.50 m s⁻¹ at 25.0° to its original direction. Calculate the speed and direction (relative to the original direction of the first puck, with angles measured on the opposite side if appropriate) of the second puck.

(a) Answer: 2.94 m s⁻¹ at 39.0° on the opposite side of the original direction

• Evidence level: Excellence (resolve momentum parallel and perpendicular to the original direction).

• Along the original direction: 1.00 × 5.00 = 1.00 × 3.50 × cos 25.0° + 0.800 v₂x, so m₂v₂x = 1.83 kg m s⁻¹.

• Perpendicular: 0 = 1.00 × 3.50 × sin 25.0° + m₂v₂y, so m₂v₂y = −1.48 kg m s⁻¹ (opposite side).

• p₂ = √(1.83² + 1.48²) = 2.35 kg m s⁻¹, so v₂ = 2.35 ÷ 0.800 = 2.94 m s⁻¹ at 39.0° on the other side of the original line.

(b) 4 small masses are held by a light frame: 2.00 kg at (1.50, 0.500) m; 0.500 kg at (3.00, 0.500) m; 1.00 kg at (2.00, 1.00) m; 3.00 kg at (0.500, 0) m. (i) Calculate the position of the centre of mass. (ii) The 2.00 kg mass is removed. Calculate the new position of the centre of mass and the distance it has moved.

(b) Answer: (i) (1.23, 0.346) m; (ii) (1.11, 0.278) m, a shift of 0.138 m

• Evidence level: Merit (take moments about the axes to find the centre of mass).

• (i) xcm = Σmx/Σm = 8.00 ÷ 6.50 = 1.23 m; ycm = Σmy/Σm = 2.25 ÷ 6.50 = 0.346 m.

• (ii) Remaining mass = 4.50 kg, moments 5.00 and 1.25, so the new centre of mass is (1.11, 0.278) m.

• Distance moved = √((1.11 − 1.23)² + (0.278 − 0.346)²) = 0.138 m.

(c) An astronaut of mass 75.0 kg is at a height of 6.37 × 10⁶ m above Earth's surface. (G = 6.67 × 10⁻¹¹ N m² kg⁻²; mass of Earth 5.97 × 10²⁴ kg; radius of Earth 6.37 × 10⁶ m.) (i) Calculate the gravitational field strength at this height and the astronaut's weight there. (ii) Explain why astronauts in a stable orbit at this height feel weightless even though gravity is still acting on them.

(c) Answer: (i) 2.45 N kg⁻¹ and 184 N; (ii) both are in free fall, so there is no contact force

• Evidence level: Excellence (use Newton's law of gravitation and explain apparent weightlessness).

• (i) g = GM/r² with r = 6.37 × 10⁶ + 6.37 × 10⁶ = 1.27 × 10⁷ m: g = 3.98 × 10¹⁴ ÷ 1.27 × 10⁷² = 2.45 N kg⁻¹.

• Weight = mg = 75.0 × 2.45 = 184 N (compare 736 N at the surface).

• Achievement: (ii) States that the astronaut and the spacecraft are both falling together.

• Merit: Links to the force: gravity is the only force acting on both, so they have the same acceleration.

• Excellence: Full explanation: gravity still acts (weight 184 N), but the astronaut and spacecraft both accelerate at gh = 2.45 m s⁻² towards Earth as they orbit; there is no contact force from the floor (no 'normal reaction'), which is what we feel as weight, so the astronaut feels weightless.