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

Exercises · 5.12

10\ \text{keV}

Andhra Pradesh BieapTextbookSubjective· 3mImportance★★★★★est

29% · 24/83 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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) …

Start your 14-day free trial to unlock the full solution →

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:

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

Start your 14-day free trial to unlock the full solution →

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

Showing the 12 most recent of 18 on this concept.

AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
$mv^2$
›Reveal solutionSolution
Kinetic energy is a scalar that depends only on speed, not direction. Since the speed stays constant at $v$ throughout the motion, the change in kinetic energy is (B) zero.

Concept and Intuition

It's tempting to think that because the velocity vector changes direction by 180° while going around half a circle, something about the energy must change too. But kinetic energy $KE=\tfrac12mv^2$ depends on the magnitude of velocity (speed), not its direction. Uniform circular motion keeps the speed fixed even as the direction continuously rotates — that's exactly why centripetal force (which is always perpendicular to velocity) does zero work.

Step-by-Step Solution

Initial kinetic energy: $KE_i=\tfrac12mv^2$ (speed is $v$ ).

After traversing half the circle, the particle's speed is still $v$ (given as constant), so $KE_f=\tfrac12mv^2$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
$t_A$
›Reveal solutionSolution
Equal kinetic energy reached under the same constant force relates time-to-reach-KE to $\sqrt{m}$ ; here $t_A:t_B = \sqrt{20:5}=2:1$ .

Concept and Intuition

Under a constant force $F$ starting from rest, acceleration $a=F/m$ and velocity at time $t$ is $v=at=\dfrac{Ft}{m}$ . Kinetic energy is $KE=\dfrac12 mv^2 = \dfrac{F^2t^2}{2m}$ . For the same $F$ and same target $KE$ , larger mass needs proportionally more time (specifically, time scales as $\sqrt{m}$ ).

Step-by-Step Solution

$KE_A = \dfrac{F^2 t_A^2}{2m_A}$ , $KE_B = \dfrac{F^2 t_B^2}{2m_B}$ .

Setting $KE_A = KE_B$ : $\dfrac{t_A^2}{m_A} = \dfrac{t_B^2}{m_B}$ .

$\dfrac{t_A^2}{t_B^2} = \dfrac{m_A}{m_B} = \dfrac{20}{5} = 4 \Rightarrow \dfrac{t_A}{t_B} = 2$ .

So $t_A : t_B = 2:1$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
$9 \times 10^{-31}$
›Reveal solutionSolution
An electron starting from rest gains kinetic energy $eV$ after falling through potential difference $V$ ; equating this to $\tfrac12mv^2$ and solving for $v$ gives $8000\ \mathrm{km\,s^{-1}}$ .

Concept and Intuition

When a charge $q$ moves through a potential difference $V$ , the electric field does work $W=qV$ on it. If the electron starts at rest and all of this work converts into kinetic energy (no other forces, non-relativistic speeds), then $eV=\tfrac12 mv^2$ . This is simply energy conservation applied to the electric force, and it's the standard way accelerating potentials (as in CRTs, electron guns, etc.) are related to the resulting electron speed.

Step-by-Step Solution

Energy gained by the electron: $eV = \tfrac12 mv^2$ .

Solve for $v$ : $v = \sqrt{\dfrac{2eV}{m}}$ .

Substitute values: $2eV = 2\times(1.6\times10^{-19})\times180 = 5.76\times10^{-17}\ \mathrm{J}$ .

Divide by mass: $\dfrac{5.76\times10^{-17}}{9\times10^{-31}} = 6.4\times10^{13}\ \mathrm{m^2\,s^{-2}}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
$20 \mathrm{~m~s^{-1}}$
›Reveal solutionSolution
Since stopping distance $s=\dfrac{u^2m}{2F}$ for a fixed retarding force $F$ , comparing A and B gives the ratio $8:9$ .

Concept and Intuition

A "retarding force" causes deceleration $a=F/m$ (Newton's second law) — for the SAME force $F$ on different masses, the heavier body decelerates less. Using $v^2=u^2-2as$ with final velocity $v=0$ (body comes to rest), the stopping distance is $s=\dfrac{u^2}{2a}=\dfrac{u^2m}{2F}$ — so the distance is proportional to (mass) $\times$ (initial speed) $^2$ , for a fixed applied force.

Step-by-Step Solution

Deceleration of a body of mass $m$ under retarding force $F$ : $a=\dfrac{F}{m}$ .

Using $v^2=u^2-2as$ with $v=0$ : $s=\dfrac{u^2}{2a}=\dfrac{u^2m}{2F}$ .

For body A: $m_A=1.5$ kg, $u_A=20$ m/s $\Rightarrow s_A=\dfrac{(20)^2(1.5)}{2F}=\dfrac{600}{2F}$ .

For body B: $m_B=3$ kg, $u_B=15$ m/s $\Rightarrow s_B=\dfrac{(15)^2(3)}{2F}=\dfrac{675}{2F}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
$1\ \mathrm{ms}^{-1}$
›Reveal solutionSolution
This is a kinetic-energy ratio problem solved by expressing both conditions ("before" and "after speeding up") as equations in the man's unknown speed $v$ , then solving algebraically.

Concept and Intuition

Kinetic energy is $\tfrac12 mv^2$ — it depends on both mass and the square of speed, so "half the mass" and "half the kinetic energy" combine multiplicatively, not additively. Setting up both given conditions as equations lets us eliminate the boy's speed and solve directly for the man's initial speed.

Step-by-Step Solution

Let the man's mass be $M$ and initial speed $v$ . The boy's mass is $M/2$ , and let his speed be $u$ .

"Man has half the KE of the boy": $\tfrac12 M v^2 = \tfrac12\left(\tfrac12 \cdot \tfrac{M}{2}u^2\right)$ . Simplify: $Mv^2 = \tfrac12\cdot\tfrac{M}{2}u^2 = \tfrac{M u^2}{4}$ , so $v^2=\dfrac{u^2}{4}$ , giving $u=2v$ .

Man speeds up by $1\ \text{ms}^{-1}$ to match the boy's KE exactly: $\tfrac12 M(v+1)^2 = \tfrac12\cdot\tfrac{M}{2}u^2 = \tfrac12\cdot\tfrac{M}{2}(2v)^2 = Mv^2$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$t_1$
›Reveal solutionSolution
Express KE purely in terms of the constant force, mass, and time, then compare the two objects. Answer: $t_2/t_1=2$ .

Concept and Intuition

Under a constant force, $v=at=(F/m)t$ , so kinetic energy grows as $KE=\dfrac{F^2t^2}{2m}$ — for a fixed force, reaching the same KE with a larger mass takes proportionally more time (since $t\propto\sqrt m$ at fixed KE and F).

Step-by-Step Solution

$v(t)=\dfrac Fm t$ , so $KE=\dfrac12 m v^2 = \dfrac12 m\left(\dfrac{F}{m}t\right)^2=\dfrac{F^2t^2}{2m}$ .

Set $KE_1=KE_2$ : $\dfrac{F^2t_1^2}{2m_1}=\dfrac{F^2t_2^2}{2m_2} \Rightarrow \dfrac{t_1^2}{m_1}=\dfrac{t_2^2}{m_2}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
$6\ \mu\mathrm{C}$
›Reveal solutionSolution
Equating the work done by the accelerating field ( $qV$ ) to the kinetic energy gained ( $\frac{1}{2}mv^2$ ) gives $v=0.6\ \text{m/s}$ .

Concept and Intuition

When a charge is "accelerated through a potential difference $V$ ", it means the electric field does work $qV$ on it, and (starting from rest) all of that work converts into kinetic energy. This is the same energy-method used for accelerating any charged particle through a potential difference, independent of the details of the field's shape.

Step-by-Step Solution

Work done on the charge = $qV = 6\times10^{-6}\ \text{C} \times 60\ \text{V} = 3.6\times10^{-4}\ \text{J}$ .

This equals the kinetic energy gained (starting from rest): $\dfrac{1}{2}mv^2 = 3.6\times10^{-4}\ \text{J}$ .

Mass $m = 2\ \text{g} = 2\times10^{-3}\ \text{kg}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
$P_L$
›Reveal solutionSolution
Tests the $P$ – $KE$ – $m$ relation $E=P^2/2m$ ; at equal KE, momentum grows with $\sqrt{m}$ , so the heavier body has the larger momentum.

Concept and Intuition

Kinetic energy and momentum are both built from mass and velocity, but they don't scale the same way with mass. Writing $KE=\tfrac12 mv^2$ in terms of momentum $P=mv$ gives $KE=\dfrac{P^2}{2m}$ , i.e. $P=\sqrt{2mE}$ . If two bodies share the same KE, the one with more mass needs more momentum to carry that same energy — a heavy, slow body can have the same energy as a light, fast one, but its momentum is larger.

Step-by-Step Solution

Let the light body have mass $m_L$ and the heavy body mass $m_H$ , with $m_H > m_L$ .

Both have the same kinetic energy $E$ , so $P_L=\sqrt{2m_L E}$ and $P_H=\sqrt{2m_H E}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
$\dfrac{r}{2}$
›Reveal solutionSolution
Closest approach distance is inversely proportional to the square of the initial speed; doubling speed quarters the closest distance.

Concept and Intuition

As charge $q$ approaches fixed charge $Q$ head-on, its kinetic energy is progressively converted into electrostatic potential energy, becoming zero (momentarily at rest) at the closest point. Energy conservation, $\frac12mv^2=\dfrac{kqQ}{r}$ , shows $r$ depends on $1/v^2$ — doubling the initial speed quadruples the initial KE, which must be absorbed by a proportionally quartered closest distance (since PE $\propto 1/r$ ).

Step-by-Step Solution

Energy conservation: $\frac12mv^2=\dfrac{kqQ}{r} \Rightarrow r=\dfrac{2kqQ}{mv^2}$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
$Q. Two cars of different masses have same kinetic energy, then (A) Heavier car has more linear momentum (B) Heavier car has less linear momentum (C) Both cars have same linear momentum (D) If same braking force applied on the two cars, the heavier car comes to rest at a shorter distance.$
›Reveal solutionSolution
Same KE but different masses means the heavier car has the larger momentum, since $p=\sqrt{2mKE}$ . Answer: (A).

Concept and Intuition

Kinetic energy and momentum are related but distinct quantities: $KE = \dfrac{p^2}{2m}$ , so $p = \sqrt{2mKE}$ . If two objects share the same KE, the one with greater mass must have greater momentum (since $p \propto \sqrt{m}$ for fixed KE).

Step-by-Step Solution

Let both cars have kinetic energy $KE$ , with masses $m_1 < m_2$ .

From $KE = \dfrac{p^2}{2m}$ , momentum is $p = \sqrt{2mKE}$ .

Since $KE$ is identical for both, $p \propto \sqrt{m}$ — so the car with larger mass ( $m_2$ ) has the larger momentum. …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
$1\ \text{ms}^{-1}$
›Reveal solutionSolution
Setting up the KE-doubling equation and solving the resulting quadratic gives the initial velocity as $(1+\sqrt2)$ ms $^{-1}$ — the only physically valid (positive) root.

Concept and Intuition

Kinetic energy scales as the square of velocity. "KE doubles when velocity increases by a fixed amount" is a quadratic condition on the original velocity — solve it like any quadratic, then discard any unphysical (negative) root.

Step-by-Step Solution

Let the initial velocity be $v$ . New velocity is $v+1$ .

Condition: $\dfrac12 m(v+1)^2 = 2\left(\dfrac12 mv^2\right)$ .

Cancel $\dfrac12 m$ : $(v+1)^2 = 2v^2$ .

Expand: $v^2+2v+1 = 2v^2 \Rightarrow v^2-2v-1=0$ .

Quadratic formula: $v = \dfrac{2\pm\sqrt{4+4}}{2} = \dfrac{2\pm2\sqrt2}{2}=1\pm\sqrt2$ . …
Start your 14-day free trial to see the full solution →
AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
$(\hat{i}+2\hat{j}-3\hat{k})$
›Reveal solutionSolution
Kinetic energy from a vector velocity uses $|\vec v|^2$ (sum of squares of components), giving 0.7 J.

Concept and Intuition

For a velocity given in component form $\vec v = v_x\hat i + v_y\hat j + v_z\hat k$ , the speed squared is $|\vec v|^2 = v_x^2+v_y^2+v_z^2$ (Pythagoras in 3D), and kinetic energy only depends on the magnitude of velocity, not direction.

Step-by-Step Solution

$m = 100\ \mathrm{g} = 0.1\ \mathrm{kg}$ .

$\vec v = \hat i + 2\hat j - 3\hat k$ m/s, so $|\vec v|^2 = 1^2+2^2+(-3)^2 = 1+4+9=14\ \mathrm{m^2/s^2}$ . …
Start your 14-day free trial to see the full solution →

←Previous question5.11 — A body constrained to move along the z-axis of a coordinate system is …→Next question5.13 — A rain drop of radius 2mm falls from a height of 500m above the ground…

🎓Unlock everything free for 14 days

Unlock everything free for 14 days