Question

If the momentum of a body is increased by 30% its kinetic energy will increase by

Answer 1

K.E = p^2/2m
So kinetic energy is directly proportional to square of momentum.
Hence,
K/K' = (p/p')^2
p' = p+ (30/100) × p
= (13p/10)
K/K'= [p/(13p/10)]^2 = 100/169
》K' = 169K/100
So, % increase in K =
{[169K/100 - K]/K} × 100
= (69/100) ×100 = 69%

Answer 2

Given: The  momentum of a body is increased by 30%

To find The kinetic energy increased

Solution: We know that kinetic energy of a body(K)=1/2mv², where m=mass of that body, v=velocity of that body

∴ K=1/2mv²

    =(1×m²v²)/2m  [multiplying numerator and denominator by 'm']

    =p²/2m     [p stands for the momentum of that body and we p=m×v

                          from Newton's second law]

Let p1 be 100kg-m/sec and is the original momentum of that body and p2 be the increased momentum. K1 be the original kinetic energy and K2 be the increased kinetic energy.

∴ p2=(100+30% of 100)

⇒p2=(100+(30/100)×100)kg-m/sec

⇒p2=(100+30)kg-m/sec

       =130 kg-m/sec

K1 =p1²/2m  and   K2= p2²/2m

K1  = (100)²/2m and K2=(130)²/2m

K1 =(100×100)/2m and K2=(130×130)/2m  

∴ The percent of change in kinetic energy

={(change in kinetic energy(K2-K1)/original momentum(K1))×100}%

={((130×130-100×100)/2m)/((100×100)/2m)}×100%

=(6900/10000)×100%  [cancelling out the 2m from numerator and denominator]

=69%

Hence the  kinetic energy increased is 69%.