Physics, asked by aviskarmishra, 8 months ago

A body of mass 5 kg is moving with velocity of 10 ms-1
. What will be the ratio of its initial kinetic
energy and final kinetic energy, if the mass of the body is doubled and its velocity is halved?

Answers

Answered by Anonymous
0

Answer:

\large\bf { Initial \ Kinetic \ energy = \frac {1}{2} mv^{2}}

\large\bf { \frac{1}{2} × 5 × 10^{2}}

\large\bf { 250}

\large\bf { final \ kinetic \ energy = ½ × 10 | 5²</p><p>= 125}

\large\bf { \frac {250}{125} = 2 }

Answered by juhijahanvi
1

Answer:

2:1

Explanation:

let, initial mass of the body ( m ) = 5 kg

initial velocity of the body ( v ) = 10 m/s

So , initial kinetic energy = 1/2 mv^2

= 1/2×5×(10)^2

= 250 J

now, m' = 2m

v' = v/2 m/s

hence, final kinetic energy will be

1/2 m'v'^2

= 1/2× 2m× (v/2)^2

= mv^2/4

put the values,

5×(10)^2/4 = 5×10×10/4

= 125J

Hence the ratio of initial kinetic energy to final kinetic energy will be 250/125 i.e., 2/1 or 2:1.

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