If the linear momentum of a moving object changes by two times, then its kinetic energy will change by a factor of
- (a)2
- (b)4
- (c)6
- (d)8
Correct — B, 4. Write kinetic energy in terms of momentum rather than speed: with p = mv, KE = ½mv² = p²/2m. Mass is unchanged, so KE depends on the square of the momentum alone. Double p and the new energy is (2p)²/2m = 4p²/2m, four times the old value. The same result comes out of the speed picture — doubling p at fixed mass doubles v, and squaring a doubled speed gives four.
- (a)2 — Two is the factor for momentum itself, not for energy. Energy carries the square, so the factor cannot be the same as the one applied to p.
- (c)6 — Six comes from no relation between p and KE. Squaring a doubling can only give four.
- (d)8 — Eight is the cube of two, which would need KE proportional to p³. The dependence is quadratic, not cubic.
Momentum is linear in speed and kinetic energy is quadratic in it, so the two never scale together. The bridge between them is KE = p²/2m, which lets any statement about momentum be converted into a statement about energy without knowing the mass or the speed.
The item is designed so that the careless reader carries the factor 2 straight across. Anyone who writes the relation down first cannot make that mistake, because the square is sitting in plain view. Note also what stays fixed — the object is the same object, so m does not change, and that is what allows p²/2m to be used as a pure proportionality.
- p = mv and KE = ½mv², so KE = p²/2m for a single object of fixed mass.
- If momentum changes by a factor k, kinetic energy changes by k²; if kinetic energy changes by a factor k, momentum changes by the square root of k.
- Two objects of different mass with equal momentum have kinetic energies in inverse ratio to their masses — the lighter body carries more energy.
- Momentum is a vector and is conserved in every collision; kinetic energy is a scalar and is conserved only in an elastic one.
Once KE is written as p²/2m the exponent does all the work — no numbers are needed.
- Carrying the factor 2 across from momentum to energy because both are called 'motion' quantities.
- Assuming the mass changes as well, which turns a clean proportionality into an unsolvable problem.
- Reading 'changes by two times' as 'increases by two units'.
Asked as a scaling question — momentum or speed is multiplied by a stated factor and you are asked what happens to kinetic energy, or the reverse.
If the linear momentum of a moving object gets doubled due to application of a force, then its kinetic energy will
- (a) remain same
- (b) increase by four times
- (c) increase by two times
- (d) increase by eight times
Answer(b) increase by four times
The identical calculation set two years earlier for the CDS, down to the doubled momentum and the fourfold energy. Both papers are testing whether you can write KE = p²/2m before you start.
- practice — not a real PYQ
If the speed of a moving body is halved, its kinetic energy becomes
- (a)half
- (b)one-fourth
- (c)double
- (d)unchanged
Answer(b) one-fourth — kinetic energy goes as the square of speed, so halving v multiplies KE by (1/2)² = 1/4.
- practice — not a real PYQ
Two bodies of masses 2 kg and 8 kg have equal momentum. The ratio of their kinetic energies (2 kg body first) is
- (a)1 : 4
- (b)4 : 1
- (c)1 : 2
- (d)2 : 1
Answer(b) 4 : 1 — with KE = p²/2m and p equal for both, kinetic energy is inversely proportional to mass, so the ratio is 8 : 2 = 4 : 1.