Question 23M.3.SL.TZ1.6
| Date | May 2023 | Marks available | [Maximum mark: 10] | Reference code | 23M.3.SL.TZ1.6 |
| Level | SL | Paper | 3 | Time zone | TZ1 |
| Command term | Calculate, Predict, Show that | Question number | 6 | Adapted from | N/A |
A uniform rod of length 1.20 m is at rest on a horizontal surface. The rod is pivoted at its centre so that it is free to rotate about a vertical axis through the centre.
A particle of mass 0.200 kg moving with speed 12.0 m s−1 collides with and sticks to one end of the rod.
The moment of inertia of the rod about the axis is 0.180 kg m2. Show that the moment of inertia of the rod–particle system is about 0.25 kg m2.
[1]
0.180 + 0.200 × 0.602 «= 0.252 kg m2» ✓
Show that the angular speed of the system immediately after the collision is about 5.7 rad s−1.
[2]
angular speed of particle = «12/0.6 = » 20 «rad s−1»
OR
angular momentum of particle «0.200 × 12.0 × 0.60» = 1.44 «Js» ✓
«angular momentum of rod-particle system 0.252 ω»
equating ω = «» = 5.71 rad s−1 ✓
For MP2, working or answer to at least 3 SF should be seen.
Calculate the energy lost during the collision.
[2]
× 0.200 × 12.02 − (0.252) × 5.712 ✓
10.3 J ✓
Award [1] for answer 11.5 J that neglects moment of inertia of particle but do not penalize this omission in (d)(i).
A frictional torque of magnitude 0.152 N m acts on the system just after it begins to rotate. Calculate
the angular deceleration of the rod.
[1]
= = 0.603 rads−2 ✓
Accept negative values.
the number of revolutions made by the rod until it stops rotating.
[2]
θ = = 27.0 rad ✓
N = = 4.3 ✓
In another situation the rod rests on a horizontal frictionless surface with no pivot. Predict, without calculation, the motion of the rod–particle system after the collision.
[2]
the rod will rotate «about centre of mass» ✓
«centre of mass» will move along straight line
«parallel to the particle’s initial velocity» ✓
For MP2, mention of translational motion is not enough.