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MHT-CET Physics · Formula sheet

Laws of Motion formulas

6 formulas and 12 common traps for MHT-CET Physics Laws of Motion, grouped by subtopic.

Full notes with worked examples

Newton's Laws: Lifts, Pulleys, Circles and Power

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Lifts, Blocks in Contact and Pulleys

Second law

∑F⃗=ma⃗,Nlift=m(g±a)\sum \vec{F} = m\vec{a}, \qquad N_{\text{lift}} = m(g \pm a)

Force, Work, Power and Circular Dynamics

Work and power

W=F⃗⋅r⃗,P=F⃗⋅v⃗,Fc=mv2rW = \vec{F}\cdot\vec{r}, \qquad P = \vec{F}\cdot\vec{v}, \qquad F_c = \frac{mv^2}{r}

Braking, Penetration and Avoiding Collision

Uniform retardation

v2=u2−2as,ssafe=(vA−vB)22av^2 = u^2 - 2as, \qquad s_{\text{safe}} = \frac{(v_A - v_B)^2}{2a}

Common traps

Subtracting a for a lift accelerating up

Accelerating UPWARD (starting up, or slowing on the way down) the floor must push harder: N = m(g + a). The reading falls only when the acceleration points down.

Using the applied force for the far block

The push on the far block is only what accelerates it: m₂F/(m₁ + m₂). 5 N on a 6 kg and a 4 kg block gives 2 N on the 4 kg block.

Taking tension at the bottom as mg

At the bottom the string must also supply the centripetal force: T = mg + mv²/L. Released from horizontal, that is 3mg.

Using average power for power at an instant

Power at an instant is F·v, and it grows as the body speeds up. A 1 kg body pushed from rest to 10 m/s in 2 s receives 25 W at t = 1 s but 50 W at t = 2 s.

Assuming speed falls linearly with distance

Speed squared falls linearly with distance. Half the speed in 30 cm leaves only a quarter of the energy, which lasts 10 cm — not another 30 cm.

Using each car's own stopping distance

Whether two cars collide depends on their RELATIVE motion: the gap must cover (v_A − v_B)²/2a, not v_A²/2a.

Impulse, Momentum and Collisions

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Impulse and Rates of Momentum

Impulse

J⃗=∫F⃗ dt=Δp⃗,F=dpdt\vec{J} = \int \vec{F}\,dt = \Delta\vec{p}, \qquad F = \frac{dp}{dt}

Collisions and the Coefficient of Restitution

Restitution

e=v2−v1u1−u2e = \frac{v_2 - v_1}{u_1 - u_2}

Common traps

Forgetting the rebound doubles the change

A ball that bounces straight back at the same speed changes momentum by 2mv, not mv. The pressure of 1000 balls a second at 50 m/s on 1 cm² is 10⁶ Pa with the factor 2.

Adding the areas of a force graph without their signs

Impulse is the SIGNED area under F–t: a force pointing backward subtracts. Only then does it equal the change in momentum.

Rebound height e·h

The rebound SPEED is e times the impact speed; height goes as speed squared, so the height is e²h.

Conserving kinetic energy in every collision

Only momentum is always conserved. Kinetic energy is conserved only when e = 1; bodies that stick lose the most.

Equilibrium and the Centre of Mass

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Moments, Reactions and the Centre of Mass

Equilibrium and CM

∑F⃗=0,∑τ=0,xcm=m1x1+m2x2m1+m2\sum\vec{F} = 0, \quad \sum\tau = 0, \qquad x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}

Common traps

Taking moments about the wrong point

Take moments about the support whose reaction you do NOT want. For the reaction at A, take moments about B: N_A·r = W(r − x).

Placing the centre of mass at the midpoint

Only for equal masses. For unequal masses it sits closer to the heavier one, still on the line joining them.

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