Question

12. Two bodies of equal masses are moving with
uniform velocities v and 2v . Find the ratio of their
kinetic energies.
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Answer 1

$\large\sf\underline{Given\::}$

• Masses of two bodies are equal

• Velocity of $\sf\:1^{st}$ body = v

• Velocity of $\sf\:2^{nd}$ body = 2v

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$\large\sf\underline{To\:find\::}$

• Ratio of kinetic energy of first and second body

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$\large\sf\underline{Concept\::}$

Here we are said that the masses of two bodies are equal, we would assume m as the masses of the two bodies . Velocity of the two bodies are given as v and 2v . We are asked to calculate the ratio of their kinetic energy. In order to do so we need to first calculate their kinetic energy using the formula and simply find their ratio . Let's begin !

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$\large\sf\underline{Formula\:to\:be\:used\::}$

• $\sf\:Kinetic\:energy\:=\:\frac{1}{2} \times mv^{2}$

where m stands for mass of the body and v stands for velocity of the body .

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$\large\sf\underline{Assumption\::}$

Let the masses of the bodies be m.

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$\large\sf\underline{Solution\::}$

Let's calculate the kinetic energy of $\sf\:1^{st}$ body :

$\sf\implies\:Kinetic\:energy_1\:=\:\frac{1}{2} \times m \times v^{2}$

$\small{\underline{\boxed{\mathrm{\implies\:Kinetic\:energy_1\:=\:\frac{mv^{2}}{2}}}}}$

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Let's calculate the kinetic energy of $\sf\:2^{nd}$ body :

$\sf\implies\:Kinetic\:energy_2\:=\:\frac{1}{2} \times m \times (2v)^{2}$

$\sf\implies\:Kinetic\:energy_2\:=\:\frac{1}{2} \times m \times 4v^{2}$

$\small{\underline{\boxed{\mathrm{\implies\:Kinetic\:energy_2\:=\:\frac{4mv^{2}}{2}}}}}$

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Now let's calculate their ratio :

$\sf\:\frac{Kinetic\:energy_1}{Kinetic\:energy_2}$

• Substituting the values we got

$\sf\to\:\frac{\frac{mv^{2}}{2}}{\frac{4mv^{2}}{2}}$

• Now when we convert division into multiplication we do know that fraction should be taken as reciprocal

$\sf\to\:\frac{m {v}^{2} }{2} \times \frac{2}{4m {v}^{2} }$

• Multiplying the terms

$\sf\to\:\frac{2m {v}^{2}}{8mv^{2}}$

• Reducing the fraction in the lower terms

$\sf\to\:\frac{\cancel{2}m {v}^{2}}{\cancel{8}mv^{2}}$

$\sf\to\:\frac{1m {v}^{2}}{4mv^{2}}$

$\sf\to\:\frac{1\cancel{m {v}^{2}}}{4\cancel{mv^{2}}}$

$\sf\to\:\frac{1}{4}$

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Therefore the required ratio is :

$\small{\underline{\boxed{\mathrm\red{\to\:\frac{Kinetic\:energy_1}{Kinetic\:energy_2}\:\:\:=\:\:\:\frac{1}{4}\:\:\:oR\:\:\:1:4}}}}$ ★

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!! Hope it helps !!

Answer 2

Answer:

1/2mv2

Explanation:

1÷2(0)(0)=o kinetics energy =0