Question

In a ballistics demonstration a police officer fires a bullet of mass 50.0g with the speed 200ms on soft plywood of thickness 2.0cm.the bullet emerges with only 10% of its initial kinetic energy.what is the emergent speed of the bullet.

Answer 1

Mass of bullet = m = 50g = 0.05kg

Initial velocity of bullet = u = 200m/s

Initial energy of the bullet, 1/2 mu square
= 1/2(0.05)(200square) = 1000J

The bullet emerges with only 10% of its KE. Let the final velocity with which it emerges to be v

So, final KE is 1/2 mv square = 10% of 1/2 mu square

= 1/2 mv square = (10/100) × 1000

= v square = 100 × 2/0.05

= v = 63.24 m/s

Hope this helps you friend
Thanks ✌️ ✌️

Answer 2

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

• 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%).