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Collision revision notes

Every idea in the Collision chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 20 and download the PDF.

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How we count: 7× asked means 7 past JEE Main questions in the PrepWiser bank needed that concept, and "last" is the latest year one appeared. ★ marks ideas asked 8 or more times.

Impulse and coefficient of restitution

Momentum and impulse (reminder)
3× asked, last 2023
J⃗=F⃗avg Δt=Δp⃗,∑mu⃗=∑mv⃗\vec J=\vec F_{\text{avg}}\,\Delta t=\Delta\vec p,\qquad \sum m\vec u=\sum m\vec v
Momentum is conserved in every collision if outside impulse is negligible; KE is conserved only in elastic ones.
  • In a split or explosion momentum is conserved but KE increases (internal energy is released). Find the unknown velocity from momentum first, then compare KE before and after.
  • Not mv/Δtmv/\Delta t for a body that rebounds. Its momentum change is 2mv2mv when it comes back at the same speed, so the average force is 2mv/Δt2mv/\Delta t.
Coefficient of restitution
5× asked, last 2023
e=speed of separationspeed of approach=v2−v1u1−u2e=\frac{\text{speed of separation}}{\text{speed of approach}}=\frac{v_2-v_1}{u_1-u_2}
Velocity components along the line of impact only. e=1e=1 elastic, 0<e<10<e<1 inelastic, e=0e=0 bodies stick.
  • Quick method. Two unknown final velocities need two equations: momentum and v2−v1=e(u1−u2)v_2-v_1=e(u_1-u_2). Both are linear, so this is faster than writing KE conservation even when e=1e=1.
Ball hitting a floor or wall
2× asked, last 2023
vn=env0,hn=e2nh,e=h1/hv_n=e^n v_0,\qquad h_n=e^{2n}h,\qquad e=\sqrt{h_1/h}
Smooth, fixed floor; dropped from rest at height hh. nn = number of bounces.
uuv′v'α\alphaβ\beta
e=0.5e=0.5, α=40∘\alpha=40^\circ: the normal part shrinks by ee, so it leaves flatter, tan⁡β=tan⁡α/e\tan\beta=\tan\alpha/e (β≈59∘\beta\approx59^\circ)
  • Hit at α\alpha, rebound at β\beta, both from the normal, on a smooth fixed wall: only the normal part reverses, times ee. So v′=usin⁡2α+e2cos⁡2αv'=u\sqrt{\sin^2\alpha+e^2\cos^2\alpha}, tan⁡β=tan⁡α/e\tan\beta=\tan\alpha/e, J=m(1+e)ucos⁡αJ=m(1+e)u\cos\alpha.
  • Not eheh or 1−e1-e. Height and KE both go as v2v^2, so the rebound height is e2he^2h and the fraction of KE lost per bounce is 1−e21-e^2 (e=1/2e=1/2 loses 75%, not 50%).
Bouncing until it stops
T=2hg  1+e1−e,S=h 1+e21−e2T=\sqrt{\frac{2h}{g}}\;\frac{1+e}{1-e},\qquad S=h\,\frac{1+e^2}{1-e^2}
Total time and total distance before the ball comes to rest; dropped from rest at height hh.

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Practise Collision

16 more concepts in 5 topics

Elastic collision in one dimension, Inelastic collisions, Bullet–block and ballistic pendulum and Oblique collisions, plus ideas from rotation and orbits. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Elastic collision in one dimension

  • Head-on elastic collision7×
  • Special cases of elastic collision4×
  • KE transferred in an elastic hit1×
  • Two blocks and a spring2×

Inelastic collisions

  • Head-on collision with restitution e
  • KE lost in a collision4×
  • Perfectly inelastic (bodies stick)6×
  • Collision during projectile motion4×

Bullet–block and ballistic pendulum

  • Ballistic pendulum3×
  • Bullet passing through or bouncing back2×
  • Embedded bullet, then friction or a fall3×
  • Speed before impact2×
  • Hit, then a full vertical circle2×
  • Also remember1

Oblique collisions

  • Momentum in two dimensions8×
  • Equal masses, elastic, target at rest2×
  • Elastic, unequal masses2×

Also asked here: rotation and orbits

  • Key points2

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About 40 past JEE Main questions from this chapter, with instant answer checking and step-by-step solutions where available. Free, and you can start without signing up.

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