The linear momentum of a body of mass m is p. Then its kinetic energy will be (A) mp (B) mp² (C) p²/m (D) p²/2

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Q. The linear momentum of a body of mass m is p. Then its kinetic energy will be (A) mp (B) mp^2 (C) p^2/m (D) p^2/2m

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Concept understanding — Kinetic Energy Speed Relation

Kinetic Energy and Speed: The Intuition

Imagine pushing a heavy box across the floor. The harder you push, the faster it moves. But here's the surprising part: doubling the speed does not require double the work — it requires four times the work. That's the core of the kinetic energy–speed relation.

Why? Because kinetic energy isn't about how fast you're moving — it's about how much effort it took to get you moving that fast. And effort (work) depends on both force and distance. When you push something to a higher speed, you have to apply force over a longer distance, and that extra distance multiplies the work required.

Note

Kinetic energy is the energy an object possesses because of its motion. A stationary object has zero kinetic energy.

The Precise Statement

The kinetic energy $K$ of an object of mass $m$ moving with speed $v$ is:

K = \frac{1}{2} m v^2

This is the kinetic energy–speed relation. The key point: kinetic energy is proportional to the square of the speed, not the speed itself.

K = \frac{1}{2} m v^2

Why the Square? A Simple Derivation

Start from Newton's second law: $F = ma$ . If a constant force $F$ acts on an object initially at rest over a displacement $s$ , the work done is $W = F s$ .

From kinematics, for constant acceleration starting from rest: $v^2 = 2 a s$ . So $s = \frac{v^2}{2a}$ .

Substitute into work:

W = F s = (ma) \left( \frac{v^2}{2a} \right) = \frac{1}{2} m v^2

That work becomes the object's kinetic energy. The square comes from the kinematic relation $v^2 = 2as$ — a direct consequence of how distance and speed are linked under constant acceleration.

Watch out

A common mistake: thinking kinetic energy is $\frac{1}{2} m v$ or $m v^2$ . The factor $\frac{1}{2}$ is essential — it comes from the integration of force over distance.

What This Means in Practice

Speed change	Kinetic energy change
Double speed ( $2v$ )	$K$ becomes $4 \times$ original
Triple speed ( $3v$ )	$K$ becomes $9 \times$ original
Halve speed ( $v/2$ )	$K$ becomes $1/4$ of original

This explains why:

A car crash at 100 km/h is four times as destructive as one at 50 km/h (four times the energy to dissipate).
Braking distance quadruples when speed doubles (because brakes must do four times the work).
A bullet at twice the speed penetrates much deeper than twice as far.

Important

Kinetic energy depends only on mass and speed — not on direction. Two objects with the same mass and speed have the same kinetic energy, even if moving in opposite directions.

Units

In SI units:

Mass $m$ in kilograms (kg) …

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In terms of momentum, $KE=p^2/2m$ .

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If the momentum of a body is increased by 100%, then the percentage increase in the kinetic energy is$
(a) 150%
(b) 200%
(c) 300%
(d) None of these
›Reveal solutionSolution

KE = p^2 / (2m), so KE scales with the SQUARE of momentum; doubling p means KE becomes 4 times as large, i.e. a 300% increase over the original.

Kinetic energy in terms of momentum: KE = p^2 / (2m)

Let the initial momentum be p, so initial KE = p^2/(2m).

Momentum increased by 100% means the new momentum is p' = 2p.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. When the momentum of body is increased by three times, its K.E. becomes:$
(a) Twice
(b) Half
(c) Four times
(d) Nine times
›Reveal solutionSolution

$KE = p^2/2m$ , so $KE \propto p^2$ — tripling $p$ makes $KE$ nine times as large.

Momentum is $p = mv$ and kinetic energy is $KE = \tfrac12 mv^2$ . Writing $KE$ in terms of $p$ : $KE = \tfrac12 mv^2 = \dfrac{(mv)^2}{2m} = \dfrac{p^2}{2m}$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If the linear momentum of the object is increased by 0.1% then the kinetic energy is increased by:$
(a) 0.4%
(b) 0.1%
(c) 0.01%
(d) 0.2%
›Reveal solutionSolution

Since KE = p^2/2m depends on the square of momentum, a small percentage change in p produces twice that percentage change in KE: 2 x 0.1% = 0.2%.

Kinetic energy can be written in terms of linear momentum p = mv as

KE = (1/2)mv^2 = p^2/(2m)

Taking logarithms to relate fractional changes (mass m is constant):

ln(KE) = 2 ln p - ln(2m)

Differentiating:

delta(KE)/KE = 2(deltap/p)

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. Kinetic energy of a body of mass 2 kg is 36 joule. Its momentum in kg m/s will be$
(a) 6
(b) 12
(c) 24
(d) 36
›Reveal solutionSolution

p = sqrt(2 m KE) = sqrt(144) = 12 kg m/s. Answer (B).

Kinetic energy in terms of momentum: KE = p^2/(2m), so p = sqrt(2 m KE).

Given m = 2 kg and KE = 36 J:

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. For constant momentum (p), the correct graph between kinetic energy (E) and mass (m) will be$
(a) decreasing hyperbola-like E-m curve
(b) straight line increasing with mass
(c) upward-increasing E-m curve
(d) none of these
›Reveal solutionSolution

With p constant, E = p^2/(2m) is proportional to 1/m -> a hyperbola-like decreasing curve. Answer (A).

Kinetic energy can be written as E = p^2/(2m).

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Kinetic energies of two moving bodies of masses 1 gm and 4 gm are equal. The ratio of their linear momentum is (A) 4 : 1 (B) sqrt(2) : 1 (C) 1 : 2 (D) 1 : 16$
›Reveal solutionSolution

With equal kinetic energies, the ratio of momenta of the two bodies is 1:2.

Kinetic energy in terms of momentum: $E_k = \dfrac{p^2}{2m} \Rightarrow p = \sqrt{2mE_k}$ .

Since both bodies have the same kinetic energy $E_k$ :

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Two bodies of masses 2 kg and 32 kg have equal kinetic energy. What is the ratio of their momentum?$
(a) 1 : 1
(b) 1 : 2
(c) 1 : 4
(d) 2 : 1
›Reveal solutionSolution

Since KE = p^2/2m, momentum for a given kinetic energy is p = sqrt(2mKE); with equal KE, the momentum ratio reduces to sqrt(m1/m2).

Kinetic energy: KE = p^2 / (2m), so p = sqrt(2m * KE).

Both bodies have the same KE (call it E), so: …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If the linear momentum of a body increases by 20%, what will be the percentage increase in K.E.?$
(a) 40%
(b) 36%
(c) 44%
(d) 20%
›Reveal solutionSolution

Since $KE = p^2/(2m)$ , kinetic energy grows with the square of momentum — so a 20% rise in momentum produces a 44% rise in kinetic energy, not 20% or 40%.

Kinetic energy expressed in terms of momentum $p$ and mass $m$ :

$KE = p^2/(2m)$

If momentum increases by 20%, the new momentum is $p' = 1.2p$ .

New kinetic energy:

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The kinetic energy of a body of mass 1 kg is 1 joule. Its velocity will be (A) 0.45 ms^-1 (B) 1 ms^-1 (C) 1.4 ms^-1 (D) 4.4 ms^-1$
›Reveal solutionSolution

$v=\sqrt{2KE/m}\approx1.4\,\text{m/s}$ .

Kinetic energy: $KE=\tfrac12mv^2$ . Rearranging, $v=\sqrt{\dfrac{2\,KE}{m}}$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The linear momentum of a body of mass m is p. Then its kinetic energy will be (A) mp (B) mp^2 (C) p^2/m (D) p^2/2m$
›Reveal solutionSolution

In terms of momentum, $KE=p^2/2m$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The momentum of any body is increased by 20%. Then its kinetic energy will increase by (A) 20% (B) 66% (C) 44% (D) 88%$
›Reveal solutionSolution

A 20% rise in momentum raises kinetic energy by 44%.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. What is the kinetic energy of a 2 kg object moving at a velocity of 3 m/s ?$
(a) 6 J
(b) 9 J
(c) 18 J
(d) 27 J
›Reveal solutionSolution

Kinetic energy of a moving object is KE = ½mv²; here it works out to 9 J.

The kinetic energy of an object of mass m moving with speed v is:

$KE = \frac{1}{2}mv^2$

Substituting m = 2 kg, v = 3 m/s:

…

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Q.The linear momentum of a body of mass m is p. Then its kinetic energy will be (A) mp (B) mp^2 (C) p^2/m (D) p^2/2m

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Q.The linear momentum of a body of mass m is p. Then its kinetic energy will be (A) mp
(B) mp^2
(C) p^2/m
(D) p^2/2m

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