Question

Solve a bullet of mass 50 gm with a speed of 200m/s on soft plywood of thickness 2cm. The bullet emerges with only 10% of its initial kinetic energy.What is the emergent speed of the bullet

Answer 1

Given that

Mass of bullet = 50 gm

Speed = 200 m/s

Thickness of plywood = 2 cm

Initial velocity of the bullet (u) = 200 m/s

To find

Emergent speed = ?

As we know,

Mass is not in its SI unit. So, we'll first of all, change its unit to SI unit.

Mass of bullet = 50 g = 0.05 kg

Secondly, we'll find the intial energy of the bullet.

Assumption

Let the final velocity of the bullet be v.

$\mathsf{\bold{Initial\:energy\:of\:the\:bullet = \frac{1}{2}mu^2}}$

$\mathsf{Initial\:energy\:of\:the\:bullet = \frac{1}{2}\:(0.05 \times 200^2)}$

$\mathsf{Initial\:energy\:of\:the\:bullet = \frac{1}{2}\:(0.05 \times 200 \: 200)}$

$\mathsf{Initial\:energy\:of\:the\:bullet = \frac{1}{2}\:(2,000)}$

$\mathsf{Initial\:energy\:of\:the\:bullet = \frac{1}{\cancel{2}} \times \cancel{2,000}}$

$\mathsf{\bold{Initial\:energy\:of\:the\:bullet = 1,000}}$

As the bullet emerges with only 10% of the bullet's kinetic energy,

Now, final kinetic energy will be,

$\mathsf{\bold{So,\:final\:kinetic\:energy\:=\:\frac{1}{2}mv^2 = 10\%of\:its\:mu^2}}$

$\mathsf{Final\:kinetic\:energy\:=\:\frac{1}{2}mv^2 = \frac{10}{100} \times 1,000}$

$\mathsf{\frac{1}{2} mv^2 = \times 100}$

$\mathsf{v^2 = 100 \times \frac{2}{0.05}}$

$\mathsf{v^2 = \frac{200}{0.05}}$

$\mathsf{v^2 = 4,000}$

$\mathsf{v = \sqrt{4,000}}$

$\mathsf{\bold{v = 63.24\:m/s\:(approximately)}}$

Answer 2

Answer:

Emergent speed of the bullet is $63.24 \: \: m {s}^{ - 1}$

Explanation:

Mass of bullet $= m = 50 \: g = 0.05 \: kg$

Initial velocity of bullet $= u = 200 \: m/s$

Initial energy of the bullet, ½ mu square =$½ (0.05)(200 )^2= 1000 \: J$

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

So, final KE is $½ mv^2 = 10%$ of ½ mu square

$= ½ mv^2= (10/100) x 1000$

$= v ^2= 100 \times ( {x}^{2} /0.05)$

= v = 63.24 m/s