Physics · JEE Advanced only

Collision of Point Masses with Rigid Bodies (JEE Advanced)

Not in the JEE Main syllabus, but it is in JEE Advanced. Study time: about 6 to 8 hours. Spend 2 hours on theory and understanding impulse-momentum relations for rigid bodies. Then 4 to 6 hours solving varied problems: rod struck at end, disc struck tangentially, and cases with different coefficients of restitution..

Last updated: 26 October 2026

Direct answer

This topic deals with what happens when a moving point mass (like a ball or bullet) collides with a rigid body that can both translate and rotate.

This topic deals with what happens when a moving point mass (like a ball or bullet) collides with a rigid body that can both translate and rotate. Unlike simple particle collisions, here you must track both linear and angular momentum, since the rigid body will start spinning after impact. The collision can be elastic, inelastic, or perfectly inelastic, and the point of impact determines how much rotation is produced.

Official syllabus

What the JEE Advanced syllabus says

“Collision of point masses with rigid bodies”

Source: JEE Advanced 2026 Information Bulletin, Annexure-I. Closest JEE Main topic: Elastic and inelastic collisions in one and two dimensions.

JEE Main only tests collisions between point masses where rotation is not involved. For Advanced, you need to handle cases where the struck object is an extended body with a moment of inertia, combining rotational dynamics with collision physics.

Key ideas

What you must understand

Linear momentum of the system is conserved if no external horizontal force acts during collision
Angular momentum about any fixed point (or about the center of mass) is conserved during collision
The coefficient of restitution relates the relative velocities at the point of contact, not at the centers of mass
Impulse acts at the point of collision and causes both linear and angular impulse on the rigid body
Energy is conserved only in perfectly elastic collisions; use the restitution equation otherwise
For a rod or disc hit at one end, the center of mass moves AND the body rotates about its center of mass

Formulas

Formulas and key facts

Linear momentum: m₁v₁ + M·V_cm = m₁v₁' + M·V_cm'
Angular momentum about point: L = I_cm·ω + M·V_cm·r
Coefficient of restitution: e = (v₂' − v₁')/(v₁ − v₂) at contact point
Velocity of contact point on rigid body: v_contact = V_cm + ω × r
Angular impulse: J·d = I_cm·Δω (where d is perpendicular distance from CM to line of impulse)
For uniform rod: I_cm = ML²/12
Impulse-momentum: J = m·Δv for point mass, J = M·ΔV_cm for rigid body

Avoid

Common mistakes

Using restitution equation with center of mass velocities instead of contact point velocities
Forgetting that the rigid body gains both translational AND rotational motion from a single impulse
Taking angular momentum about a moving point without accounting for the motion of that point
Ignoring the moment of inertia and treating the rigid body as a point mass

Worked example

A JEE Advanced-level example

Question. A uniform rod of mass M and length L lies on a smooth horizontal surface. A point mass m moving with velocity v₀ perpendicular to the rod strikes one end and sticks to it. Find the angular velocity of the system after collision and the velocity of the center of mass of the system.

Show solution

Since no external horizontal force acts, linear momentum is conserved: m·v₀ = (m + M)·V_cm. So V_cm = m·v₀/(m + M).
Since no external torque acts about the center of mass of the system, angular momentum about the system's CM is conserved. First, find the new CM position from the rod's center: x = m·(L/2)/(m + M) toward the struck end.
Initial angular momentum about system CM: L_i = m·v₀·(L/2 − x) = m·v₀·(L/2)·M/(m + M) = m·M·v₀·L / [2(m + M)].
Moment of inertia of system about new CM: I = ML²/12 + M·x² + m·(L/2 − x)². After simplification: I = ML²/12 + mM·L²/[4(m + M)] = ML²(m + M/3)/[4(m + M)/3] = ML²(M + 4m)/[12(m + M)] + mML²/[4(m+M)]. Simplified: I = (ML²/12)·(1 + 3m/(m+M)) = ML²(M + 4m)/[12(m + M)].
Using L_i = I·ω: ω = [m·M·v₀·L / (2(m + M))] / [ML²(M + 4m)/(12(m + M))] = 6m·v₀ / [L(M + 4m)].
Final answers: V_cm = m·v₀/(m + M) and ω = 6m·v₀/[L(M + 4m)].

Answer: Velocity of CM = mv₀/(m + M); Angular velocity ω = 6mv₀/[L(M + 4m)]

Solved twice independently; both solutions agree.

FAQs

Frequently asked questions

Is Collision of Point Masses with Rigid Bodies in the JEE Main syllabus?

No. It is in the JEE Advanced 2026 syllabus only. The closest JEE Main topic is Elastic and inelastic collisions in one and two dimensions.

How long does Collision of Point Masses with Rigid Bodies take to prepare for JEE Advanced?

About 6 to 8 hours. Spend 2 hours on theory and understanding impulse-momentum relations for rigid bodies. Then 4 to 6 hours solving varied problems: rod struck at end, disc struck tangentially, and cases with different coefficients of restitution., if you already know the related JEE Main topic.