Question

how much will the momentum increase if the kinetic energy of an object is increased by 50%?​

Answer 1

Answer:

Percentage increase in K.E,

(E)=[

E

1

E

2

−E

1

]×100=[

E

1

E

2

−1]×100

But Eαp

2

⇒

E

1

E

2

=

p

1

2

p

2

2

∴ % increase in K.E. =[

p

1

2

p

2

2

−1]×100

Let p

1

=100, then

p

2

=100+50=150% increase in

K.E.=[

(100)

2

(150)

2

−1]×100=[

100

225

−1]×100

=125%

Therefore, the percentage increase in its kinetic energy is 125%

Answer 2

First, derive the relation between momentum and kinetic energy.

$E_{K} = \frac{1}{2}m {v}^{2}\\ E_{K} = \frac{1}{2}m {v}^{2} \times \frac{m}{m} \\ E_{K} = \frac{ {m}^{2} {v}^{2} }{2m} \\ E_{K} = \frac{ {p}^{2} }{2m} \: \: \: (eqn \: 1)$

New Kinetic Energy = 150% of E.k = 3E.k/2

Let new momentum be p'.

New relation is

$\frac{3}{2} E_{K} = \frac{ {p'}^{2} }{2m} \: \: \: (eqn \: 2)$

Divide eqn 2 by eqn 1.

$\frac{\frac{3}{2} E_{K}}{E_{K} } = \frac{\frac{ {p'}^{2} }{2m}}{\frac{ {p}^{2} }{2m}} \\ \frac{3}{2} = \frac{{p'}^{2}}{ {p}^{2} }$

Now calculate % increase in p.