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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%

302

Mukasa

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