Q.A mass of 5 kg is moving along a circular path of radius 1 m. If the mass moves with 300 revolutions per minute, its kinetic energy would be

Showing the 12 most recent of 18 on this concept.

1. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If a particle of mass 'm' covers half of the horizontal circle with constant speed 'v', then the change in its kinetic energy is (A) $mv^2$ (B) zero (C) $2mv^2$ (D) $\dfrac{1}{2}mv^2$

   ›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

   1. Initial kinetic energy: $KE_i=\tfrac12mv^2$ (speed is $v$).
   2. After traversing half the circle, the particle's speed is still $v$ (given as constant), so $KE_f=\tfrac12mv^2$. …

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2. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.Two bodies A and B of masses 20 kg and 5 kg respectively are at rest. Due to the action of a force of 40 N separately, if the two bodies acquire equal kinetic energies in times $t_A$ and $t_B$ respectively, then $t_A : t_B =$ (A) 1 : 2 (B) 2 : 1 (C) 2 : 5 (D) 5 : 6

   ›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

   1. $KE_A = \dfrac{F^2 t_A^2}{2m_A}$, $KE_B = \dfrac{F^2 t_B^2}{2m_B}$.
   2. Setting $KE_A = KE_B$: $\dfrac{t_A^2}{m_A} = \dfrac{t_B^2}{m_B}$.
   3. $\dfrac{t_A^2}{t_B^2} = \dfrac{m_A}{m_B} = \dfrac{20}{5} = 4 \Rightarrow \dfrac{t_A}{t_B} = 2$.
   4. So $t_A : t_B = 2:1$. …

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3. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.The velocity acquired by an electron at rest when subjected to a uniform electric field of potential difference 180 V is (Mass of electron = $9 \times 10^{-31}$ kg and charge of electron = $1.6 \times 10^{-19}$ C) (A) $400\ \mathrm{km\,s^{-1}}$ (B) $4000\ \mathrm{km\,s^{-1}}$ (C) $800\ \mathrm{km\,s^{-1}}$ (D) $8000\ \mathrm{km\,s^{-1}}$

   ›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

   1. Energy gained by the electron: $eV = \tfrac12 mv^2$.
   2. Solve for $v$: $v = \sqrt{\dfrac{2eV}{m}}$.
   3. Substitute values: $2eV = 2\times(1.6\times10^{-19})\times180 = 5.76\times10^{-17}\ \mathrm{J}$.
   4. Divide by mass: $\dfrac{5.76\times10^{-17}}{9\times10^{-31}} = 6.4\times10^{13}\ \mathrm{m^2\,s^{-2}}$. …

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4. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.Two bodies A and B of masses 1.5 kg and 3 kg are moving with velocities $20 \mathrm{~m~s^{-1}}$ and $15 \mathrm{~m~s^{-1}}$ respectively. If the same retarding force is applied on the two bodies, then the ratio of the distances travelled by the bodies A and B before they come to rest is (A) $1:1$ (B) $8:9$ (C) $2:3$ (D) $3:8$

   ›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

   1. Deceleration of a body of mass $m$ under retarding force $F$: $a=\dfrac{F}{m}$.
   2. Using $v^2=u^2-2as$ with $v=0$: $s=\dfrac{u^2}{2a}=\dfrac{u^2m}{2F}$.
   3. For body A: $m_A=1.5$ kg, $u_A=20$ m/s $\Rightarrow s_A=\dfrac{(20)^2(1.5)}{2F}=\dfrac{600}{2F}$.
   4. For body B: $m_B=3$ kg, $u_B=15$ m/s $\Rightarrow s_B=\dfrac{(15)^2(3)}{2F}=\dfrac{675}{2F}$. …

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5. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.A running man has half kinetic energy to that of a boy of half of his mass. The man speeds up by $1\ \mathrm{ms}^{-1}$, so as to have same kinetic energy as that of the boy. The initial speed of the man is (A) $\sqrt{2}\ \mathrm{ms}^{-1}$ (B) $(\sqrt{2}-1)\ \mathrm{ms}^{-1}$ (C) $\dfrac{1}{(\sqrt{2}-1)}\ \mathrm{ms}^{-1}$ (D) $\dfrac{1}{\sqrt{2}}\ \mathrm{ms}^{-1}$

   ›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

   1. Let the man's mass be $M$ and initial speed $v$. The boy's mass is $M/2$, and let his speed be $u$.
   2. "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 $u=2v$.
   3. Man speeds up by $1\ \text{ms}^{-1}$ to match the boy's KE exactly: $\tfrac12 M(v+1)^2 = \tfrac12\cdot\tfrac{M}{2}u^2 = \tfrac12\cdot\tfrac{M}{2}(2v)^2 = Mv^2$. …

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6. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.Two objects having masses 1:4 ratio are at rest. When both of them are subjected to same force separately, they achieved same kinetic energy during times $t_1$ and $t_2$ respectively. Then ratio of $\dfrac{t_2}{t_1}$ is (A) $4$ (B) $2$ (C) $2.5$ (D) $1$

   ›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

   1. $v(t)=\dfrac Fm t$, so $KE=\dfrac12 m v^2 = \dfrac12 m\left(\dfrac{F}{m}t\right)^2=\dfrac{F^2t^2}{2m}$.
   2. Set $KE_1=KE_2$: $\dfrac{F^2t_1^2}{2m_1}=\dfrac{F^2t_2^2}{2m_2} \Rightarrow \dfrac{t_1^2}{m_1}=\dfrac{t_2^2}{m_2}$. …

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7. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.A particle of mass 2 g and charge $6\ \mu\mathrm{C}$ is accelerated from rest through a potential difference of 60 V. The speed acquired by the particle is (A) $0.6\ \mathrm{ms^{-1}}$ (B) $1.2\ \mathrm{ms^{-1}}$ (C) $1.8\ \mathrm{ms^{-1}}$ (D) $0.3\ \mathrm{ms^{-1}}$

   ›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

   1. Work done on the charge = $qV = 6\times10^{-6}\ \text{C} \times 60\ \text{V} = 3.6\times10^{-4}\ \text{J}$.
   2. This equals the kinetic energy gained (starting from rest): $\dfrac{1}{2}mv^2 = 3.6\times10^{-4}\ \text{J}$.
   3. Mass $m = 2\ \text{g} = 2\times10^{-3}\ \text{kg}$. …

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8. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.A light body of momentum '$P_L$' and a heavy body of momentum '$P_H$', both have the same kinetic energy, then (A) $P_L > P_H$ (B) $P_H > P_L$ (C) $P_L = P_H$ (D) Always $P_H = 2P_L$

   ›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

   1. Let the light body have mass $m_L$ and the heavy body mass $m_H$, with $m_H > m_L$.
   2. Both have the same kinetic energy $E$, so $P_L=\sqrt{2m_L E}$ and $P_H=\sqrt{2m_H E}$. …

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9. AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
   Q.A particle of charge q is shot with speed 'V' towards another fixed particle of charge Q. It reaches Q upto a closest distance 'r' and then returns. If q were shot with speed 2V then closest distance of approach of Q is (A) r (B) 2r (C) $\dfrac{r}{2}$ (D) $\dfrac{r}{4}$

   ›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

   1. Energy conservation: $\frac12mv^2=\dfrac{kqQ}{r} \Rightarrow r=\dfrac{2kqQ}{mv^2}$. …

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10. 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

    1. Let both cars have kinetic energy $KE$, with masses $m_1 < m_2$.
    2. From $KE = \dfrac{p^2}{2m}$, momentum is $p = \sqrt{2mKE}$.
    3. Since $KE$ is identical for both, $p \propto \sqrt{m}$ — so the car with larger mass ($m_2$) has the larger momentum. …

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11. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
    Q.The kinetic energy of a car is doubled when its velocity is increased by $1\ \text{ms}^{-1}$. Then the initial velocity of the car is (A) $(2+\sqrt{2})\ \text{ms}^{-1}$ (B) $(1-\sqrt{2})\ \text{ms}^{-1}$ (C) $(2-\sqrt{2})\ \text{ms}^{-1}$ (D) $(1+\sqrt{2})\ \text{ms}^{-1}$

    ›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

    1. Let the initial velocity be $v$. New velocity is $v+1$.
    2. Condition: $\dfrac12 m(v+1)^2 = 2\left(\dfrac12 mv^2\right)$.
    3. Cancel $\dfrac12 m$: $(v+1)^2 = 2v^2$.
    4. Expand: $v^2+2v+1 = 2v^2 \Rightarrow v^2-2v-1=0$.
    5. Quadratic formula: $v = \dfrac{2\pm\sqrt{4+4}}{2} = \dfrac{2\pm2\sqrt2}{2}=1\pm\sqrt2$. …

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12. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.A toy car of mass 100 g is moving with velocity of $(\hat{i}+2\hat{j}-3\hat{k})$ m, then the kinetic energy of the car is (A) 7 J (B) 70 J (C) 0.7 J (D) 0.07 J

    ›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

    1. $m = 100\ \mathrm{g} = 0.1\ \mathrm{kg}$.
    2. $\vec v = \hat i + 2\hat j - 3\hat k$ m/s, so $|\vec v|^2 = 1^2+2^2+(-3)^2 = 1+4+9=14\ \mathrm{m^2/s^2}$. …

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