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Push a ruler lying on a smooth table exactly at its centre and it slides without turning. Push it near one end and it slides and spins. JEE Advanced questions on "collision of point masses with rigid bodies" all build on this: one impulse, two separate effects. This post shows how to split them, works a free-rod example and a hinged-rod example with numbers, and explains the point you can strike without jarring the pivot.
What does an impulse do to a rigid body?
An impulse J is a large force acting for a short time. On a rigid body it does two independent things:
Linear: J = MΔv_cm. The centre of mass responds as if the whole mass were there, wherever you strike.
Angular: Jd = I_cmΔω, where d is the perpendicular distance from the centre of mass to the line of the blow. This is the short-time form of NCERT's dL/dt = τ_ext in its rotational motion chapter.
The JEE (Advanced) 2026 brochure lists impulse, angular momentum and "collision of point masses with rigid bodies". Check your exam year's bulletin.
Keep the two equations apart. Close and Heron (2011) found that many students assumed linear and angular momentum are conserved jointly, and few applied l = r × p correctly. Their tutorial helped students see that the two are conserved separately.
Worked example 1: a free rod struck at one end
A uniform rod of mass 1.2 kg and length 1.0 m lies on a frictionless table. A blow of 0.60 N s is applied at one end, perpendicular to the rod.
I_cm = ML²/12 = 1.2 × 1.0/12 = 0.10 kg m².
Centre of mass: v_cm = J/M = 0.60/1.2 = 0.50 m/s.
Spin: ω = Jd/I_cm = 0.60 × 0.50/0.10 = 3.0 rad/s.
Struck end: 0.50 + 3.0 × 0.50 = 2.0 m/s forward.
Far end: 0.50 − 3.0 × 0.50 = −1.0 m/s, so it moves backwards.
Point at rest: where ωr = v_cm, so r = 0.50/3.0 ≈ 0.17 m from the centre, on the far side.
Kinetic energy: ½ × 1.2 × 0.50² + ½ × 0.10 × 3.0² = 0.15 + 0.45 = 0.60 J. Check: the work done by an impulse is J times the average of the struck point's speed before and after, 0.60 × (0 + 2.0)/2 = 0.60 J. The same blow at the centre gives only 0.15 J, because there is no spin.
Where can you strike without jarring the pivot?
For a free rod struck at distance d from the centre, the point momentarily at rest is at r = I_cm/(Md) = L²/(12d) on the other side. If a pivot sits exactly at that point, it needs to supply no impulse. For a pivot at one end, r = L/2, so d = L/6, which puts the strike 2L/3 from the pivot. This point is called the centre of percussion. For the 1.0 m rod it is about 0.67 m from the pivot.
Worked example 2: a rod hinged at its top
Hang the same rod from a frictionless hinge at its top end and strike it horizontally with 0.60 N s. Now the hinge can push back, so linear momentum is not conserved. Use angular impulse about the hinge, where I = ML²/3 = 0.40 kg m².
Struck at the bottom (1.0 m from the hinge): ω = 0.60 × 1.0/0.40 = 1.5 rad/s, v_cm = 1.5 × 0.50 = 0.75 m/s. Linear: 0.60 + J_hinge = 1.2 × 0.75 = 0.90, so the hinge gives 0.30 N s in the same direction as the blow.
Struck at 2L/3 (0.67 m): ω = 0.60 × (2/3)/0.40 = 1.0 rad/s, v_cm = 0.50 m/s, M v_cm = 0.60 N s. The hinge impulse is zero.
Struck at the centre (0.50 m): ω = 0.75 rad/s, v_cm = 0.375 m/s, M v_cm = 0.45 N s, so the hinge gives 0.15 N s backwards.
That sign change is a classic MCQ. For a bullet that embeds in a hinged rod, see our post on conservation of angular momentum, and for the linear version with a hanging block, the ballistic pendulum.
Which mistakes cost marks?
Using J = MΔv_cm on a hinged body. The hinge impulse must be included.
Taking torques about the struck point. Use the centre of mass for a free body, the hinge for a pivoted one.
Using I_cm about the hinge, or ML²/3 for a free rod.
Assuming kinetic energy is conserved in a sticking collision.
A practice routine
Repeat worked example 1 for a blow at L/4 from the centre and find the point at rest.
Derive the 2L/3 result from r = I_cm/(Md) without notes.
Do worked example 2 for a disc pivoted on its rim.
Finish one problem with the energy check J × (average speed of struck point).
If you are unsure which axis to take moments about, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Lay a ruler on a smooth table, flick it at different points, and predict which point stays still before you look.
