Question

Linear momentum of particle is increased by (a) 100% (b) 1% without changing its mass. Find percentage increase in its kinetic energy in both cases.

Answer 1

Answer:

300%

Explanation:

KE is proportional to p^2 so if

p is doubled then KE becomes 4 times

means increase of 3 times means 300%

Answer 2

answer : (a) 300% (b) 2%

(a) linear momentum of particle is increased by 100 %.

let initial linear momentum = P

final linear momentum = P + 100% P = 2P

relation between kinetic energy and linear momentum.

K.E = P²/2m ⇒K.E ∝ P²

so, K.E₁/K.E₂ = P₁²/P₂²

⇒K.E/K.E₂ = P²/(2P)² = 1/4

⇒K.E₂ = 4P

so increase in kinetic energy = (K.E₂- K.E₁)/K.E₁ × 100

= (4P - P)/P × 100

= 300 %

( b) linear momentum of particle is increased by 1% without changing its mass.

for below 10% we have to use approximation method.

here K.E ∝ P²

so, ∆K.E/K.E = 2 × ∆P/P

⇒ % increase in kinetic energy = 2 × % increase in linear momentum.

⇒% increase in kinetic energy = 2 × 1% = 2%