An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy $10\ \text{keV}$, and the second with $100\ \text{keV}$. Which is faster, the electron or the proton? Obtain the ratio of their speeds. (electron mass $= 9.11 \times 10^{-31}\ \text{kg}$, proton mass $= 1.67 \times 10^{-27}\ \text{kg}$, $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$).

> [!TLDR] > The electron is faster. Using the kinetic energy relation $K = \frac12 m v^2$, the speed ratio is $\frac{v_e}{v_p} = \sqrt{\frac{m_p K_e}{m_e K_p}} \approx 13.5$, so the electron moves about 13.5 times faster than the proton. The core idea here is simple: kinetic energy depends on both mass and speed. When two particles have different masses but comparable kinetic energies, the lighter one must be moving much faster. This is a direct consequence of $K = \frac12 m v^2$. Let’s work through it step by step. 1. **Write the kinetic energy relation for each particle.** For the electron: $K_e = \frac12 m_e v_e^2$ For the proton: $K_p = \frac12 m_p v_p^2$ 2. **We want the ratio of speeds $v_e / v_p$.** Divide the two equations: \[ \frac{K_e}{K_p} = \frac{\frac12 m_e v_e^2}{\frac12 m_p v_p^2} = \frac{m_e v_e^2}{m_p v_p^2} \] 3. **Rearrange to isolate the speed ratio.** \[ \frac{v_e^2}{v_p^2} = \frac{K_e}{K_p} \cdot \frac{m_p}{m_e} \] Taking square roots: \[ \frac{v_e}{v_p} = \sqrt{ \frac{K_e}{K_p} \cdot \frac{m_p}{m_e} } \] 4. **Plug in the numbers.** $K_e = 10\ \text{keV}$, $K_p = 100\ \text{keV}$, so $K_e / K_p = 0.1$ $m_e = 9.11 \times 10^{-31}\ \text{kg}$, $m_p = 1.67 \times 10^{-27}\ \text{kg}$ \[ \frac{m_p}{m_e} = \frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}} \approx 1833 \] Therefore: \[ \frac{v_e}{v_p} = \sqrt{0.1 \times 1833} = \sqrt{183.3} \approx 13.54 \] …

An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10keV, and the

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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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The electron is faster. Using the kinetic energy relation $K = \frac12 m v^2$ , the speed ratio is $\frac{v_e}{v_p} = \sqrt{\frac{m_p K_e}{m_e K_p}} \approx 13.5$ , so the electron moves about 13.5 times faster than the proton.

The core idea here is simple: kinetic energy depends on both mass and speed. When two particles have different masses but comparable kinetic energies, the lighter one must be moving much faster. This is a direct consequence of $K = \frac12 m v^2$ .

Let’s work through it step by step.

Write the kinetic energy relation for each particle.

For the electron: $K_e = \frac12 m_e v_e^2$

For the proton: $K_p = \frac12 m_p v_p^2$
We want the ratio of speeds $v_e / v_p$ .

Divide the two equations:

\frac{K_e}{K_p} = \frac{\frac12 m_e v_e^2}{\frac12 m_p v_p^2} = \frac{m_e v_e^2}{m_p v_p^2}

Rearrange to isolate the speed ratio.

\frac{v_e^2}{v_p^2} = \frac{K_e}{K_p} \cdot \frac{m_p}{m_e}

Taking square roots:

\frac{v_e}{v_p} = \sqrt{ \frac{K_e}{K_p} \cdot \frac{m_p}{m_e} }

Plug in the numbers. $K_e = 10\ \text{keV}$ , $K_p = 100\ \text{keV}$ , so $K_e / K_p = 0.1$ $m_e = 9.11 \times 10^{-31}\ \text{kg}$ , $m_p = 1.67 \times 10^{-27}\ \text{kg}$

\frac{m_p}{m_e} = \frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}} \approx 1833

Therefore:

$\frac{v_e}{v_p} = \sqrt{0.1 \times 1833} = \sqrt{183.3} \approx 13.54$ …

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

KCET 2026Set C21 markMCQ
$Q. Two bodies with kinetic energies in the ratio of 3:1 are moving with equal linear momentum. The ratio of their masses is (A) 1:4 (B) 1:3 (C) 1:2 (D) 1:1$
›Reveal solutionSolution

For a given (fixed) linear momentum, kinetic energy $KE=p^2/2m$ is inversely proportional to mass — this lets a kinetic-energy ratio be converted directly into a mass ratio.

Step 1 — Express KE in terms of momentum and mass
$KE = \frac{p^2}{2m} \implies m = \frac{p^2}{2\,KE}$
Since both bodies have the same momentum $p$ , $m \propto \dfrac{1}{KE}$ .

Step 2 — Apply the given kinetic-energy ratio

Given $KE_1:KE_2 = 3:1$ : …
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COMEDK 2025Set 2025-E1 markMCQ
$300 \%$
›Reveal solutionSolution
Kinetic energy is proportional to the square of momentum, so tripling momentum multiplies kinetic energy by 9, an increase of 800%.

Concept & Intuition

The relationship between kinetic energy $K$ and momentum $p$ is direct and quadratic:
$K = \frac{p^2}{2m}$
Since mass $m$ is constant, any change in momentum causes the kinetic energy to change by the square of the momentum multiplier. If momentum triples, kinetic energy becomes $3^2 = 9$ times larger. The percentage change is then $(9 - 1) \times 100\% = 800\%$ . The trap is to think the percentage increase is 300% (just the momentum increase) or 900% (confusing the factor with the percentage).

Step-by-step reasoning

Write the original kinetic energy Let the original momentum be $p$ . Then the original kinetic energy is

$K_1 = \frac{p^2}{2m}$

Write the new kinetic energy after tripling momentum New momentum $p' = 3p$ . So

$K_2 = \frac{(3p)^2}{2m} = \frac{9p^2}{2m}$

Find the ratio of new to original kinetic energy

$\frac{K_2}{K_1} = \frac{\frac{9p^2}{2m}}{\frac{p^2}{2m}} = 9$
So $K_2 = 9K_1$ .

Calculate the percentage change …
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KCET 2024Set D-21 markMCQ
$_0^1 n + _{92}^{235}U \rightarrow _{56}^{144}Ba + _{36}^{89}Kr + 3_0^1 n$
›Reveal solutionSolution
$Q$ = (total BE of products) − (total BE of reactants); divide the 180 MeV released among the 3 neutrons.

Step 1 — The concept: energy released = gain in binding energy

Binding energy is the energy that would be needed to pull a nucleus apart into free nucleons. A nucleus with a larger binding energy is more tightly bound, i.e. lower in energy. So when nucleons rearrange into more tightly-bound products, the surplus binding energy is released:
$Q = \sum BE(\text{products}) - \sum BE(\text{reactants})$
Step 2 — Tally the binding energies

The reaction is
$_0^1 n + \;_{92}^{235}\mathrm{U} \;\longrightarrow\; _{56}^{144}\mathrm{Ba} + \;_{36}^{89}\mathrm{Kr} + 3\,_0^1 n$
A free neutron is a single nucleon, so it has zero binding energy — neither the incident neutron nor the three emitted neutrons contribute to either side of the tally. (Check the bookkeeping first: nucleons $1+235=236$ and $144+89+3=236$ ✓; charge $92 = 56+36$ ✓.)

Reactants: $BE(^{235}\mathrm{U}) = 1800$ MeV (the incident neutron adds 0).

Products: $BE(^{144}\mathrm{Ba}) + BE(^{89}\mathrm{Kr}) = 1200 + 780 = 1980$ MeV (the 3 neutrons add 0).

Step 3 — Compute $Q$
$Q = 1980 - 1800 = 180\ \text{MeV}$
This is the total energy released by the fission — comfortably in line with the real figure of $\sim200$ MeV per $^{235}$ U fission.

Step 4 — Share it among the neutrons …
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KCET 2023Set A-31 markMCQ
$220$
›Reveal solutionSolution

Conserve momentum and energy: the alpha carries the fraction $(A-4)/A$ of the Q value.

Step 1 — Set up conservation laws.

The parent nucleus ( $A=220$ ) is at rest, so total initial momentum is zero. After the decay we have an alpha particle (mass number $4$ ) and a daughter nucleus (mass number $A-4=216$ ). Momentum conservation requires them to fly apart with equal and opposite momenta:
$p_\alpha = p_d = p.$
The Q value is shared as kinetic energy:
$Q = K_\alpha + K_d = 5.5\ \text{MeV}.$
Step 2 — Relate the kinetic energies.

Using $K=\dfrac{p^{2}}{2m}$ with the same $p$ for both fragments,

$\frac{K_\alpha}{K_d} = \frac{m_d}{m_\alpha} = \frac{216}{4}.$ …
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COMEDK 2022Set 20221 markMCQ
$Q. If kinetic energy of a body is increased by 300%, then percentage change in momentum will be (A) 100% (B) 150% (C) 265% (D) 73.2%$
›Reveal solutionSolution

So momentum doubles: percentage change = (2p − p)/p × 100 = 100%.

Concept: KE = p²/2m → p ∝ √(KE) at fixed mass.

KE increased by 300% → KE_final = KE + 3KE = 4 KE.

p_final/p_initial = √(KE_final/KE_initial) = √4 = 2. …
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COMEDK 2021Set 20211 markMCQ
$Q. Two masses of 1g and 9g are moving with equal kinetic energy. The ratio of magnitude of their momentum is (A) 3 : 1 (B) 1 : 3 (C) 1 : 2 (D) 2 : 1$
›Reveal solutionSolution

With equal kinetic energies K: p1/p2 = sqrt(m1/m2) = sqrt(1 g / 9 g) = sqrt(1/9) = 1/3.

Concept: relation between momentum and kinetic energy, K = p^2/(2m) => p = sqrt(2 m K).

With equal kinetic energies K: …
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COMEDK 2021Set 2021-B1 markMCQ
$Q. If momentum is increased by 10% then the increase in the K.E. of the body is (A) 40% (B) 100% (C) 21% (D) 20%$
›Reveal solutionSolution

Since $KE = p^2/2m$ , kinetic energy scales as the square of momentum. Raising $p$ by 10% multiplies $KE$ by $1.1^2 = 1.21$ , i.e. a 21% increase.

For a body of mass $m$ , kinetic energy in terms of momentum is
$KE = \frac{p^2}{2m}.$
If momentum increases by 10%, then $p' = 1.10\,p$ , so …
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KCET 2020Set A-11 markMCQ
$F_n$
›Reveal solutionSolution
At a separation of 10 nm, the electromagnetic force between two protons dominates overwhelmingly because the nuclear force is a short-range force that essentially vanishes beyond about 1–2 fm (femtometres). The correct option is (B).

The key to this question is understanding the range of the two fundamental forces involved. The electromagnetic force between two charged particles follows an inverse-square law and has infinite range — it falls off as $1/r^2$ , but never truly disappears. The nuclear force (the strong nuclear force), on the other hand, is a short-range force. It binds protons and neutrons together inside the nucleus, but its influence drops to essentially zero once the separation exceeds a few femtometres ( $1 \text{ fm} = 10^{-15} \text{ m}$ ).

Here, the separation given is 10 nm = $10 \times 10^{-9} \text{ m} = 10^{-8} \text{ m}$ . Compare that to the typical range of the nuclear force, which is about $1\text{–}2 \text{ fm} = 10^{-15} \text{ m}$ . The separation is ten million times larger than the range of the nuclear force. At such a distance, the nuclear force between the two protons is effectively zero.

Let’s work through the numbers to see just how huge the difference is.

Calculate the electromagnetic force $F_e$ . Both protons have charge $e = 1.6 \times 10^{-19} \text{ C}$ . The separation is $r = 10 \text{ nm} = 10^{-8} \text{ m}$ . Coulomb’s law gives:

$F_e = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2}$
Using $\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9 \text{ N m}^2/\text{C}^2$ :
$F_e = 9 \times 10^9 \times \frac{(1.6 \times 10^{-19})^2}{(10^{-8})^2}$ $= 9 \times 10^9 \times \frac{2.56 \times 10^{-38}}{10^{-16}}$ $= 9 \times 10^9 \times 2.56 \times 10^{-22}$ $= 2.304 \times 10^{-12} \text{ N}$

Estimate the nuclear force $F_n$ at this separation. …
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