Physics, asked by Armanalitf, 1 year ago

A bullet of 80g hit a target with a velocity of 100cm/s.After hitting the target it's velocity decreases to 50cm/s.How much kinetic energy lossed?

Answers

Answered by Anonymous
4

\huge{\underline{\underline{\sf{Answer \colon}}}}

To find the net loss in kinetic energy of the bullet » ∆K

From the Question,

  • Mass of the Bullet,m = 80g

  • Initial Velocity,u = 100cm/s

  • Final Velocity,v = 50cm/s

As the units of all the quantities are in CGS system,we don't need to convert them

We know that,

\sf{\Delta{K} = {K}_{f} \ - \ {K}_{i}} \\ \\ \implies \ \sf{\Delta{K} = \frac{1}{2}m{v}^{2} - \frac{1}{2}mu{u}^{2}} \\ \\ \implies \ \boxed{\Delta{K} = \sf{\frac{m({v}^{2}-{u}^{2})}{2}}}

Putting the values,we get:

\sf{\Delta{K} = \frac{80(100+50)(50-100)}{2}} \\ \\ \implies \ \sf{\Delta{K} = 40 \times (-50) \times 150} \\ \\ \implies \ \huge{\sf{\Delta{K}= - 3,00,000 Erg}}

3,00,000 Ergs of Kinetic Energy is lost

Answered by ShivamKashyap08
7

\huge{\bold{\underline{\underline{....Answer....}}}}

\huge{\bold{\underline{Given:-}}}

m = 80 grams.

u = 100 cm/s (When it hits the target)

v = 50 cm/s (When it leaves the target)

\huge{\bold{\underline{Explanation:-}}}

\large{\bold{K.E_i = \frac{1}{2} mu^2}}

\large{\bold{K.E_f = \frac{1}{2} mv^2}}

Change in Kinetic energy (or) Kinetic energy lost.

\large{\bold{ \Delta K.E = K.E_f - K.E_i}}

\large{ \Delta K.E = \frac{1}{2} m(v^2 - u^2)}

Substituting the values

\large{ \Delta K.E = \frac{1}{2} \times 80 \times (50^2 - 100^2)}

\large{ \Delta K.E = \frac{1}{ \cancel{2}} \times { \cancel{80}} \times (- 7500)}

\large{ \Delta K.E = 40 \times - 7500}

\large{ \Delta K.E = - 3,00,000 \: erg}

Taking magnitude.

\huge{\boxed{\boxed{ \Delta K.E =  3,00,000 \: erg}}}

So,the 3,00,000 erg is lost .

If it asks in SI system then,

\large{\bold{ \star \: 1 \:erg = 10^{-7} \:J}}

\large{ \star \: 1 \:J = 10^{7} \: erg}

\large{3,00,000 \: erg = 3,00,000 \times10^{-7} J}

\large{\bold{ \therefore 3,00,000 \: erg = 3 \times 10^{-2} J}}

\huge{\boxed{\boxed{ \Delta K.E =  0.03 \: J}}}

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