Question

A bullet leaving the muzzle of a rifle barrel with a velocity v penetrates a plank and loses one fifth of its velocity. It then strikes second plank, which it just penetrates through. Find the ratio of the thickness of the planks supposing average resistance to the penetration is same in both the cases.​

Answer 1

Question:

A bullet leaving the muzzle of a rifle barrel with a velocity v penetrates a plank and loses one fifth of its velocity. It then strikes second plank, which it just penetrates through. Find the ratio of the thickness of the planks supposing average resistance to the penetration is same in both the cases.

Given:

Bullet of velocity v penetrates two planks.

Let the thickness of the first plank be d1 and the thickness of other plank be d2.

Let the velocity at which the bullet strikes the first plank is v1 and velocity for the second plank is v2

As per the question, the bullet loses one fifth of the velocity after penetrating the first plank.

To know:

• $KE= \frac{1}{2} mv^2$
• Work = Force x displacement.

Solution:

Since, the the bullet loses one fifth of the velocity, the velocity of the bullet after penetration will be:

$velocity = \bigg(v- \frac{v}{5} \bigg) \implies \frac{4v}{5}$

Now, Let's assume the resistance force is F for both the planks.

And as the bullet hits the plank, it undergoes a change in Kinetic energy and change in Kinetic energy is referred as work.

$Work = \bigg|K_{final} - K_{initial} \bigg|\\\\ F \times d = |K_{final} - K_{initial}|$

For the first plank,

$Work = \bigg| \frac{1}{2} m \bigg(\frac{4v}{5}\bigg)^2 - \frac{1}{2}mv^2 \bigg| \\\\ F \times d_1 = \frac{1}{2}mv^2 \bigg| \bigg(\frac{4}{5}\bigg)^2 - 1 \bigg| \\\\ F \times d_1 = \frac{1}{2}mv^2 \bigg| \frac{16}{25} - 1 \bigg| \\\\ F \times d_1 = \frac{1}{2}mv^2 \bigg| \frac{-9}{25} \bigg| \\\\ F \times d_1 = \frac{1}{2}mv^2 \bigg(\frac{9}{25}\bigg) - - - - - [i]$

For the second plank,

For the second plank, it is given that "it just penetrates through" means the final velocity becomes 0 and the bullet loses all its kinetic energy after the penetration.

$Work = \frac{1}{2} m \bigg(\frac{4v}{5}\bigg)^2 - 0 \\\\ F \times d_2 = \frac{1}{2} m \bigg(\frac{4v}{5}\bigg)^2 \\\\ F \times d_2 = \frac{1}{2} mv^2 \bigg(\frac{16}{25}\bigg) - - - - - [ii]$

Dividing [i] and [ii], we have,

$\frac{F \times d_1}{F \times d_2} = \frac{\frac{1}{2}mv^2 \bigg(\frac{9}{25}\bigg)} {\frac{1}{2} mv^2 \bigg(\frac{16}{25}\bigg)} \\\\ \frac{ \cancel{F} \times d_1}{\cancel{F} \times d_2} = \frac{\cancel{\frac{1}{2}mv^2} \bigg(\frac{9}{25}\bigg)} {\cancel{\frac{1}{2} mv^2} \bigg(\frac{16}{25}\bigg)} \\\\ \frac{d_1}{d_2} = \frac{9}{16}$

Hence, the ratio of the thickness of the planks is 9/16.

Answer 2

Solution:-Let the thickness of the first barrel be S1 and thickness of second barrel be S2

As It is said the resistance force is same ,so let it F for both.

Law of conservation of energy says that,work done is equal to change in Kinetic engery

W=Ki-Kf

Ki=initial kinetic energy,Kf=final kinetic energy

now,

also W=Force×distance,so

for first barrel

initial velocity=v

final=v-v/5=4v/5

W=F×S1=1/2 ×m ×16v^2/25-mv^2/2

F×S1=mv^2/2(16/25 -1)

F×S1=mv^2/2×(-9/25)/50

=-9mv^2/50. ..i)

for second barrel

initial velocity=4v/5

final velocity=0 (it stops after penetrating)

F×S2=Kf-Ki

=>1/2×m×(0)^2-1/2×m×16v^2/25=-16mv^2/50. ii)

from first and second equation

F×S1/F×S2=9/16

S1/S2=9/16

Ratio of thickness of plank=9:16