An object of mass 2000 g possesses 100 J kinetic energy. The object must be moving with a speed of
- (a)10.0 m/s
- (b)11.1 m/s
- (c)11.2 m/s
- (d)12.1 m/s
Correct — A, 10.0 m/s. Kinetic energy KE = ½mv². First convert the mass to SI units: 2000 g = 2 kg. Then 100 = ½ × 2 × v², which gives 100 = v², so v = 10 m/s. The mass must be in kilograms and the energy in joules for the answer to come out in metres per second.
- (b)11.1 m/s — Does not satisfy ½ × 2 × v² = 100; it is a near-miss value designed to reward a rounding or arithmetic slip.
- (c)11.2 m/s — Fails the check ½mv² = 100 J with m = 2 kg (which needs v = 10 m/s exactly); a distractor mimicking √125-type errors.
- (d)12.1 m/s — Also fails ½mv² = 100 J for m = 2 kg; it is a tempting but incorrect value from mishandling the ½ factor or the gram-to-kilogram conversion.
The kinetic energy of a moving body is KE = ½mv², where m is mass in kilograms and v is speed in metres per second, giving energy in joules. Rearranging, the speed is v = √(2·KE/m). Consistent SI units are essential — grams must be converted to kilograms.
The single most common error is leaving the mass as 2000 (grams) or forgetting the factor of ½. With m = 2 kg and KE = 100 J the arithmetic collapses neatly to v² = 100, so v = 10 m/s; the other options are deliberately close, non-round numbers to catch careless conversion.
- Kinetic energy KE = ½mv²; rearranged, v = √(2·KE/m).
- 1 joule = 1 kg·m²/s², so mass must be in kg and speed in m/s.
- 2000 g = 2 kg — the required unit conversion.
- Here v = √(2 × 100 / 2) = √100 = 10 m/s.
- Not converting 2000 g to 2 kg before substituting.
- Dropping the factor of ½ in KE = ½mv².
Asked as a direct substitution problem: given mass and kinetic energy, find the speed using v = √(2·KE/m).
The planet Mercury is revolving in an elliptical orbit around the sun as shown in the given figure. The kinetic energy of Mercury is greatest at the point labelled
- (a) A
- (b) B
- (c) C
- (d) D
Answer(a) A
Same concept of kinetic energy as ½mv²: Mercury moves fastest (and so has the greatest KE) at the point nearest the Sun, applying the very speed–energy relationship used to solve this numerical.
A thin disc and a thin ring, both have mass M and radius R. Both rotate about axes through their center of mass and are perpendicular to their surfaces at the same angular velocity. Which of the following is true?
- (a) The ring has higher kinetic energy
- (b) The disc has higher kinetic energy
- (c) The ring and the disc have the same kinetic energy
- (d) Kinetic energies of both the bodies are zero since they are not in linear motion
Answer(a) The ring has higher kinetic energy
Both questions turn on comparing or computing kinetic energy from the defining formula; here the linear KE = ½mv², there the rotational KE = ½Iω², the same energy concept applied to motion.
- practice — not a real PYQ
A body of mass 4 kg moves with a speed of 5 m/s. Its kinetic energy is
- (a)20 J
- (b)50 J
- (c)100 J
- (d)200 J
Answer(b) 50 J — KE = ½ × 4 × 5² = ½ × 4 × 25 = 50 J.
- practice — not a real PYQ
If the speed of a moving body is doubled, its kinetic energy becomes
- (a)unchanged
- (b)doubled
- (c)three times
- (d)four times
Answer(d) four times — kinetic energy is proportional to the square of the speed.