Question

In a ballistics demonstration, a police officer fires a bullet of mass 50.0 g with speed 200 ms-1 on soft plywood of thickness 2.00cm. The bullet emerges with only 10% of its initial kinetic energy. What is the emerge
speed of the bullet?
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Plzz yrr guysss solve it ​

Answer 1

Answer:

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$\large\underline{\tt Ur\:Answer\:is\:63.2ms-1}$

Initial Kinetic Energy :-

$\large{\tt Ki = \frac{1}{2}×\frac{50}{1000}×200×200 J = 1000 J}$

Final Kinetic Energy :-

$\large{\tt Kf =\frac{10}{100}×1000=100 J}$

If Vf is emergent speed of the bullet ,then :-

$\large{\tt \frac{1}{2}×\frac{50}{1000}×{Vf}^{2}=100}$

$\large{\tt {Vf}^{2}=4000}$

$\huge\blue{\boxed{\tt Vf=63.2ms-1}}</p><p>$

Note :- that the speed is reduced by approximately 68% and not 90%

Hope this will help uh!!

☺️❣️

Answer 2

${\huge{\underline{\underline{\mathfrak{\green{♡ANSWER♡}}}}}}$.

The initial kinetic energy of the bullet is ${mv}^2/2=1000J$

It has a final kinetic energy of${0.1×1000=100.J}$.

If ${vf}$ is the emergent speed of the bullet,

$\frac{1}{2}$ ${mv}^2$ ${f}$=${100J}$

${vf}$= $\sqrt{2×100J 0.05kg}$

=${63.2. m{s}^-1}$

☄.The speed is reduces by approximately 68% (not 90%).