Question

the kinetic energy of any object is changes 2.5% find the percentage change of linear momentum

Answer 1

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Answer 2

We know that,

Kinetic Energy (K.E) = (1/2)mV²

$K.E = (1/2)*m^{2} V^{2}/m\\  K.E = P^{2} /2m$

Where, m = mass of the object

            V = velocity of the object

            P = momentum of the object

But mass of an object is a constant. Therefore,

           K.E ∝ P²----- (1)

If kinetic energy of the object increases in 2.5%,

Let K.E1 = initial kinetic energy = 100E

     K.E2 = final kinetic energy = 102.5E

        P1 =  initial momentum

        P2 =  final momentum

For initial position,

     K.E1 ∝ P1²

    100E ∝ P1²---- (2)

For initial position,

     K.E2 ∝ P2²

     102.5E ∝ P2²---- (3)

dividing equation (3) by equation (2),

   $102.5E/100E = P2^{2} /P1^{2} \\ P2^{2} =1.025P1^{2}\\ P2 = 1.012P1$

Then we can write

Percentage change of momentum= (difference of momentum/Initial momentum)*100

                                                          =[ (P2 - P1)/P1]*100%

                                                          =[(1.012P1-P1)/P1]*100%

                                                          =[(0.012P1)/P1]*100%

                                                          = 0.012*100 = 1.2%

Answer: Percentage change of momentum = 1.2%