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Laws of Motion formulas

Class 11 physics formula sheet for NEET and JEE: the key equations of NCERT chapter 4, the special cases questions are built on, and diagrams where they help.

22 formulas5 sectionsClass 11 · Chapter 42 of 5 sections free

By Sreeraj P, M.Sc Physics · 10+ years teaching NEET and JEE

Most used formulasOther formulas and cases

1Newton's laws and momentum

  • 1st law: inertia (of rest, motion, direction); mass measures inertia. Coin tossed in a train: falls behind if train accelerates, ahead if it retards, back into hand at constant velocity.
  • 3rd law: action and reaction act on different bodies, never cancel; not valid for pseudo forces.
$$\vec F=\frac{d\vec p}{dt}=m\vec a+\vec v\frac{dm}{dt},\qquad \vec J=\int\vec F\,dt=\Delta\vec p$$
$$\begin{array}{l}\displaystyle \vec F=\frac{d\vec p}{dt}=m\vec a+\vec v\frac{dm}{dt}\\[6pt]\displaystyle \vec J=\int\vec F\,dt=\Delta\vec p\end{array}$$

Impulse = area under $F$–$t$ graph. Catching with hands lowered (more time) → smaller force. 1 kg wt $=9.8$ N; 1 N $=10^5$ dyne.

Change in momentum$|\Delta\vec p|$
Hits wall and stops / rebounds with same speed$mv$ / $2mv$ (normal to wall)
Rebounds with $v_2$ after hitting at $v_1$$m(v_1+v_2)$
Strikes at angle $\theta$ with wall, same speed$2mv\sin\theta$ ($2mv\cos\theta$ if $\theta$ from normal)
Uniform circular motion through angle $\theta$$2mv\sin\frac\theta2$ (towards centre)
Projectile: launch → top / launch → landing$mu\sin\theta$ / $2mu\sin\theta$ (downward)
Variable mass / streamForce
Machine gun, $n$ bullets of mass $m$ in time $t$$F=\dfrac{nmv}t$
Bullets holding a plate up: stop / bounce back$\dfrac{nmv}t=Mg$ / $\dfrac{2nmv}t=Mg$
Liquid jet (area $A$, density $\rho$) on wall: stops / rebounds$\rho Av^2$ / $2\rho Av^2$; rebounds at $v'$: $\rho Av(v+v')$; at angle $\theta$ to wall: $2\rho Av^2\sin\theta$
Flow through a 90° pipe bend$\sqrt2\rho Av^2$
Sand dropped on conveyor belt moving at $u$$F=u\dfrac{dm}{dt}$
Falling chain on table ($y$ fallen)force on table $=\dfrac{3Mgy}L$
Rocketthrust $=u\dfrac{dm}{dt}$; $v=u\ln\dfrac{m_0}{m}-gt$
$$\sum\vec F_{\text{ext}}=0\ \Rightarrow\ \vec p=\text{constant}$$

Recoil: $V=\dfrac{mv}M$. Bullet embeds in block: $V=\dfrac{mv}{M+m}$. Man walks $s$ on a boat: boat moves $\dfrac{ms}{M+m}$. Shell at rest breaks into two: $\dfrac{v_1}{v_2}=\dfrac{m_2}{m_1}$; into three: third momentum balances the other two. Shell exploding at top of projectile path: use $Mu\cos\theta=m_1v_1+m_2v_2$ (horizontal); one piece retraces path → other gets $\dfrac{(M+m_1)u\cos\theta}{m_2}$.

2Equilibrium, tension and pulleys

$$F=mg\tan\theta,\qquad T=\sqrt{F^2+(mg)^2}$$

Bob held at angle $\theta$ by a horizontal force. Chain hanging between walls at angle $\theta$: end tension $\dfrac{mg}{2\sin\theta}$, middle $\dfrac{mg}2\cot\theta$. Heavy rope (mass $m$) holding $M$: tension at distance $x$ from bottom $\left(M+\dfrac{mx}L\right)g$. Straightening a rope with a weight needs infinite tension.

$$a=\frac F{m_1+m_2+m_3},\qquad T_{12}=(m_2+m_3)a,\quad T_{23}=m_3a$$
$$\begin{array}{l}\displaystyle a=\frac F{m_1+m_2+m_3}\\[6pt]\displaystyle T_{12}=(m_2+m_3)a,\quad T_{23}=m_3a\end{array}$$

Blocks pulled on a smooth surface (same for contact forces when pushed). Vertical chain pulled up by $F$: $T=\dfrac{m_{\text{below}}}{m_{\text{total}}}F$.

m₁m₂Atwoodm₁m₂table + hangingm₁m₂θincline + hanging
System (smooth, light pulley)$a$$T$
Atwood ($m_1>m_2$)$\dfrac{(m_1-m_2)g}{m_1+m_2}$$\dfrac{2m_1m_2g}{m_1+m_2}$; thrust on pulley $2T$
$m_1$ on table, $m_2$ hanging$\dfrac{m_2g}{m_1+m_2}$$\dfrac{m_1m_2g}{m_1+m_2}$
… with friction $\mu$ on table$\dfrac{(m_2-\mu m_1)g}{m_1+m_2}$$\dfrac{m_1m_2g(1+\mu)}{m_1+m_2}$
$m_1$ on incline $\theta$, $m_2$ hanging$\dfrac{(m_2-m_1\sin\theta)g}{m_1+m_2}$$\dfrac{m_1m_2g(1+\sin\theta)}{m_1+m_2}$
Double incline $\alpha$, $\beta$$\dfrac{(m_2\sin\beta-m_1\sin\alpha)g}{m_1+m_2}$$\dfrac{m_1m_2g(\sin\alpha+\sin\beta)}{m_1+m_2}$
Three blocks $M_1$, $M_2$ (table), $M_3$$\dfrac{(M_1-M_3)g}{M_1+M_2+M_3}$—
Pulley accelerating up at $a_0$replace $g$ by $g+a_0$
  • Constraints: total string length constant. Movable pulley holding $m_2$, $m_1$ on the other end: $a_1=2a_2$; hanging $m_2$: $a_2=\dfrac{(m_2-2m_1)g}{m_2+4m_1}$; $m_1$ on a smooth table: $a_2=\dfrac{m_2g}{4m_1+m_2}$. Force $F$ on a light pulley whose string pulls a block: $T=F/2$, block $a=\dfrac F{2m}$, pulley $\dfrac F{4m}$.
  • Rope climbing: $T=m(g+a)$. Man of mass $M$ in a box $m$ pulling a rope over a pulley to hold it still: scale reads $\dfrac{(M-m)g}2$.

3 more sections and 14 formulas in the full chapter

  1. 3Frames and apparent weight2 formulas · 1 case table
  2. 4Friction5 formulas · 3 case tables
  3. 5Circular motion dynamics7 formulas · 1 diagram

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