If the linear momentum of the object is increased by 0.1% then the kinetic energy is increased by: (a) 0.4% (b

Question 57 of 59

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%

(d) 0.2%

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2026MCQ· 1mImportance★★★★★

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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.