Question

If the ratio of two masses are 5:7 and the ratio of velocities is 2:3. Then, find the ratio of kinetic energy​

Answer 1

Required Answer:-

Given:

• The ratio of two masses is 5 : 7.
• The ratio of their velocity is 2 : 3

To Find:

• The ratio of kinetic energy.

Solution:

→ Let the masses of these two bodies be 5x and 7x.

→ Let the velocity of these two bodies be 2y and 3y.

We know that,

→ Kinetic Energy = ½mv²

where m is the mass of the body and v is the velocity of the body.

So, Ratio of their Kinetic energy will be,

= (Kinetic Energy of first body)/(Kinetic energy of second body)

= [1/2 × 5x × (2y)²] ÷ [1/2 × 7x × (3y)²]

= 5/7 × 4y²/9y²

= 5/7 × 4/9

= 20/63

= 20 : 63

→ Therefore, the ratio of their kinetic energy will be 20 : 63

Answer:

• The ratio of kinetic energy will be 20 : 63.

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Answer 2

Given,

• The ratio of two masses are 5:7.
• The ratio of velocities is 2:3.

To Find,

• The ratio of kinetic energy.

Solution,

Let's

The m of Object A = 5X

So,

The M of Object B = 7X

Let's,

The v of Object A = 2X

So,

The V of Object B = 3X

$Kinetic \: \: \: energy = \frac{1}{2} m {v}^{2}$

The Ratio Of There Kinetic energy

$:\implies \frac{ \frac{1}{2} \times m \times {v}^{2} }{ \frac{1}{2} \times {M} \times {V}^{2} } \\ \\ :\implies \frac{ \frac{1}{2} \times m \times {v}^{2} }{ \frac{1}{2} \times {M} \times {V}^{2} } \\ \\ :\implies \frac{ 5x \times {(2x)}^{2} }{ {7x} \times {(3x)}^{2} } \\ \\ :\implies \frac{5x \times </strong><strong>4</strong><strong> {x}^{2} }{7x \times 9 {x}^{2} } \\ \\ :\implies \frac{</strong><strong>2</strong><strong>0 {x}^{3} }{63 {x}^{3} } \\ \\ :\implies \frac{</strong><strong>2</strong><strong>0}{63} \\ \\ </strong><strong>2</strong><strong>0 : 63$

Required Answer,

The ratio of The Objects kinetic energy is 20 : 63