A-Level Impulse and the conservation of momentum and energy: worked solution

Question

After an elastic collision a 0.20 kg bob on a 1.5 m string leaves at 2.857 m/s and swings up. Find the greatest angle from the vertical through which it swings.

Immediately after the collision the bob has kinetic energy only. At the highest point of its swing it is momentarily at rest, so all of that kinetic energy has become gravitational potential energy. The string is always perpendicular to the motion, so it does no work and energy is conserved. (1/2)mv² = mgh The mass m appears on both sides and cancels, so you do not need the 0.20 kg: (1/2)v² = gh h = v²/(2g) h = 2.857²/(2 × 9.81) h = 8.162/19.62 h = 0.416 m Now turn that rise into an angle. If the string of length L makes an angle theta with the vertical, the bob hangs a vertical distance L cos(theta) below the pivot, whereas at rest it hangs L below it. So the height it has risen is h = L - L cos(theta) = L(1 - cos(theta)) 0.416 = 1.5(1 - cos(theta)) 1 - cos(theta) = 0.416/1.5 1 - cos(theta) = 0.2773 cos(theta) = 0.7227 theta = cos⁻¹(0.7227) theta = 43.7 deg