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

Showing the 12 most recent of 28 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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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9. 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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10. 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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11. 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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12. 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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