The problems are on the centre of mass worksheet. Try them on your own first, then compare with the answers below.
Quick answer: The centre of mass is the mass-weighted average position, xcm = (1/M)∫x dm. These worked answers correct the 20 problems of the AP Physics C worksheet on centre of mass, from momentum of a cricket ball to billiard balls at an angle: each solution shows the formula, the substitution and the final answer with its unit.
Every answer states the principle first, then the substitution and the result with units. For the integration problems each step is shown so you can compare your own working. Take g = 9.81 m/s² and give results to two or three significant figures.
Level: AP Physics C (Mechanics and E&M, calculus-based) · Curriculum: AP PC:M Unit 4
Problem 1 Momentum of a cricket ball
1. p = mv = 0.16 × 35 = 5.6 kg·m/s.2. The momentum has the same direction as the velocity, towards the batter.
Problem 2 Impulse of a constant force
1. J = FΔt = 50 × 0.20 = 10 N·s.2. Δv = J/m = 10/2.0 = 5.0 m/s.
Problem 3 Two-mass centre
xcm = (2.0 × 0 + 6.0 × 4.0)/(2.0 + 6.0) = 24/8.0 = 3.0 m, closer to the heavier mass.Problem 4 Units of impulse
1. N·s = (kg·m/s²) × s = kg·m/s.2. N·s is natural when force and time are given (impulse); kg·m/s when mass and velocity are given (momentum).
Problem 5 Skaters pushing apart
1. Total momentum is zero: 50 × 2.8 − 70v = 0, so v = 140/70 = 2.0 m/s.2. He moves in the opposite direction to Wanjiru.
Problem 6 Egg and pillow
The change of momentum is the same, so the impulse is the same. The pillow lengthens the stopping time, and since F = J/Δt the average force is smaller.Problem 7 Three-mass centre
1. xcm = (0 + 6.0 + 0)/6.0 = 1.0 m.2. ycm = (0 + 0 + 12)/6.0 = 2.0 m.
Problem 8 Non-uniform rod
1. M = ∫₀² 3.0x dx = 1.5x² = 6.0 kg.2. ∫xλ dx = ∫₀² 3.0x² dx = x³ = 8.0; xcm = 8.0/6.0 = 1.3 m (2L/3).
Problem 9 Impulse from a graph
1. Triangle: ½ × 0.015 × 400 = 3.0 N·s.2. v = J/m = 3.0/0.060 = 50 m/s.
Problem 10 Cars locking together
1. 1200 × 15 = 2000v, so v = 9.0 m/s.2. Ki = ½ × 1200 × 225 = 1.35 × 10⁵ J; Kf = ½ × 2000 × 81 = 8.1 × 10⁴ J; loss = 54 kJ.
Problem 11 Elastic collision on a track
v₁ = (m₁ − m₂)v/(m₁ + m₂) = (0.20/0.40) × 4.0 = 2.0 m/s; v₂ = 2m₁v/(m₁ + m₂) = (0.60/0.40) × 4.0 = 6.0 m/s. Check: momentum 1.2 → 0.6 + 0.6 ✓.Problem 12 Exploding shell
1. 4.0 × 20 = 3.0v, so v = 27 m/s.2. At the same point where the unbroken shell would have landed, because gravity is the only external force.
Problem 13 Impulse mistake
1. Force × distance is work (energy), not impulse. Impulse needs force × time.2. J = 800 × 0.0040 = 3.2 N·s.
Problem 14 Time-varying force
1. J = ∫₀² 6.0t² dt = 2.0t³ = 16 N·s.2. v = J/m = 16/4.0 = 4.0 m/s.
Problem 15 Semicircular plate
1. The strip has width 2√(R² − y²) and thickness dy, so dm = 2σ√(R² − y²) dy.2. M = σπR²/2. ∫y dm = 2σ∫₀^R y√(R² − y²) dy = 2σ × R³/3 = (2/3)σR³, so ycm = (2σR³/3)/(σπR²/2) = 4R/(3π).
3. ycm = 4 × 0.30/(3π) = 0.13 m.
Problem 16 Density rising linearly
1. M = λ₀∫₀^L (1 + x/L) dx = λ₀(L + L/2) = 1.5λ₀L. ∫xλ dx = λ₀(L²/2 + L²/3) = 5λ₀L²/6. Ratio: (5/6)/(3/2) L = 5L/9.2. xcm = 5 × 0.90/9 = 0.50 m.
Problem 17 Rocket burn
1. Δv = ve ln(m₀/m) = 2000 × ln(500/150) = 2000 × 1.204 = 2.4 × 10³ m/s.2. Thrust = ve|dm/dt| = 2000 × 5.0 = 1.0 × 10⁴ N.
Problem 18 Sand on a belt
1. F = v dm/dt = 0.50 × 8.0 = 4.0 N.2. P = Fv = 4.0 × 0.50 = 2.0 W.
3. dK/dt = ½(dm/dt)v² = ½ × 8.0 × 0.25 = 1.0 W. The other 1.0 W becomes thermal energy as the sand slips against the belt.
Problem 19 Ballistic pendulum
1. Energy conservation: v = √(2gh) = √(2 × 9.81 × 0.12) = 1.53 m/s.2. Momentum: 0.010v₀ = 2.010 × 1.534, so v₀ = 3.1 × 10² m/s.
Problem 20 Billiard balls at an angle
1. x: 4.0 = 3.2 × 0.80 + vx, so vx = 1.44 m/s. y: 0 = 3.2 × 0.60 + vy, so vy = −1.92 m/s.2. v = √(1.44² + 1.92²) = 2.4 m/s, at 53° below the original direction.
3. Ki = ½ × 0.17 × 16 = 1.36 J; Kf = ½ × 0.17 × (10.24 + 5.76) = 1.36 J, so the collision was elastic.
