Elastic & Inelastic Collisions
Centre of Mass & Collisions
42
JEE Qs
28%
Hard
105
min
Always begin collision problems by identifying the type of collision (elastic, inelastic, perfectly inelastic) and applying conservation of linear momentum; then, selectively apply kinetic energy conservation or the coefficient of restitution as appropriate.
🧮 Key Formulas
✅ Key Points for JEE
- 1Linear momentum of the system of colliding bodies is *always* conserved in the absence of external forces, regardless of the type of collision.
- 2Kinetic energy is conserved *only* in elastic collisions (e=1); it is converted to other forms (heat, sound, deformation) in inelastic (0 < e < 1) and perfectly inelastic (e=0) collisions.
- 3The coefficient of restitution 'e' is a crucial parameter defining the type of collision, relating the magnitudes of relative velocities before and after impact.
- 4For 2D collisions, apply the conservation of linear momentum by resolving it into independent perpendicular components (e.g., x and y axes).
- 5Understanding the relative velocities and correctly applying sign conventions for directions of motion is critical for using the 'e' formula.
⚠️ Common Mistakes
- ✕Assuming kinetic energy is conserved in *all* collisions, rather than only in elastic collisions.
- ✕Incorrectly applying the sign convention for velocities in the coefficient of restitution formula, leading to errors in relative velocities.
- ✕Failing to resolve momentum into components for 2D collisions and incorrectly trying to apply scalar equations.
- ✕Confusing the center of mass velocity with the velocities of individual particles.
📝 Practice Questions
See allQ29.As shown below, bob A of a pendulum having massless string of length ' R ' is released from 60∘ to the vertical. It hits another bob B of half the mass that is at rest on a friction less table in the center. Assuming elastic collision, the magnitude of the velocity of bob A after the collision will be (take g as acceleration due to gravity.) (1) 4 3 √Rg (2) 23 √Rg (3) √Rg (4) 13 √Rg
Q26.Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Three identical spheres of same mass undergo one dimensional motion as shown in figure with initial velocities vA = 5 m/s, vB = 2 m/s, vC = 4 m/s. If we wait sufficiently long for elastic collision to happen, then vA = 4 m/s, vB = 2 m/s, vC = 5 m/s will be the final velocities. Reason (R): In an elastic collision ∣∣ ∣ ∣ ∣ 2025 (29 Jan Shift 2) JEE Main Previous Year Paper between identical masses, two objects exchange their velocities. In the light of the above statements, choose the correct answer from the options given below : (1) (A) is false but (R) is true (2) Both (A) and (R) are true but (R) is NOT the correct explanation of (A) (3) Both (A) and (R) are true and (R) is the correct (4) (A) is true but (R) is false explanation of (A)
Q32.Consider a circular disc of radius 20 cm with centre located at the origin. A circular hole of radius 5 cm is cut from this disc in such a way that the edge of the hole touches the edge of the disc. The distance of centre of mass of residual or remaining disc from the origin will be (1) 2.0 cm (2) 1.5 cm (3) 1.0 cm (4) 0.5 cm
Q42.The center of mass of a thin rectangular plate (fig - x ) with sides of length a and b, whose mass per unit area (σ) varies as σ = σ0xab (where σ0 is a constant), would be (1) ( 23 a, 2b ) (2) ( a2 , 2b ) (3) ( 13 a, 2b ) (4) ( 23 a, 23 b)
Q4. A stationary particle breaks into two parts of masses mA and mB which move with velocities vA and vB respectively. The ratio of their kinetic energies (KB : KA) is : (1) vB : vA (2) mB : mA (3) mBvB : mAvA (4) 1 : 1
Q22.A uniform thin metal plate of mass 10 kg with dimensions is shown. The ratio of x and y coordinates of center of mass of plate in n . The value of n is ________ 9
NCERT Chapters
- Class 11 Physics Ch 5: Laws of Motion
- Class 11 Physics Ch 6: Work, Energy and Power
- Class 11 Physics Ch 7: System of Particles and Rotational Motion