Question

18. A bullet of mass 20 g is fired from a rifle with a velocity of 800 ms–1. After passing through a mud wall 100 cm thick, velocity drops to 100 m s–1. What is the average resistance of the wall ?
(Neglect friction due to air and work of gravity)
(A) 7000 N (B) 6300 N
(C) 0 (D) 6000 N

Answer 1

Given :-

▪ A bullet of mass m = 20 g or 0.02 kg is fired from a rifle with a velocity of u = 800 m/s.

▪ After passing through a mud wall 100 cm or 1 m thick, velocity drops to v = 100 m/s

To Find :-

▪ Average resistance of the wall.

Solution :-

Given to us that the velocity of the bullet changed from 800 m/s to 100 m/s after travelling a distance of 100 cm or 1 m

So, Let us find the acceleration of the bullet while going through the wall, Using third Equation of motion, we have

⇒ 2as = v² - u²

⇒ 2×a×1 = (100)² - (800)²

⇒ 2a = 10000 - 640000

⇒ 2a = -630000

⇒ a = -315000 m/s

So, The acceleration provided by the wall is -315000 m/s , here the negative sign indicates that the acceleration is negative and hence the velocity decreased.

Now, Let us find the force provided by the wall against the wall,

⇒ Force = Mass × Acceleration

⇒ Force = 0.02 × -315000

⇒ Force = -6300 newton

Here, The negative sign indicates that the force acted against the motion of the bullet. but the magnitude or average resistance provided by the wall is 6300 newton.

Hence, Option(B) is correct.

Answer 2

GIVEN :-

• A ʙᴜʟʟᴇᴛ ᴏғ ᴍᴀss 20g ɪs ғɪʀᴇᴅ ғʀᴏᴍ ᴀ ʀɪғʟᴇ ᴡɪᴛʜ ᴀ ᴠᴇʟᴏᴄɪᴛʏ ᴏғ 800 m/s .

• Aғᴛᴇʀ ᴘᴀssɪɴɢ ᴛʜʀᴏᴜɢʜ ᴀ ᴍᴜᴅ ᴡᴀʟʟ 100ᴄm ᴛʜɪᴄᴋ, ᴠᴇʟᴏᴄɪᴛʏ ᴅᴇᴏᴘs ᴛᴏ 100 m/s .

TO FIND :-

1. Tʜᴇ ᴀᴠᴇʀᴀɢᴇ ʀᴇsɪsᴛᴀɴᴄᴇ ᴏғ ᴛʜᴇ ᴡᴀʟʟ .

SOLUTION :-

☯︎ ᴍᴀss ᴏғ ʙᴜʟʟᴇᴛ (m) = 20g = 0.02 kg

☯︎ Vᴇʟᴏᴄɪᴛʏ ᴏғ ʙᴜʟʟᴇᴛ ʙᴇғᴏʀᴇ ᴘᴀssɪɴɢ ᴛʜʀᴏᴜɢʜ ᴍᴜᴅ ᴡᴀʟʟ (ᴜ) = 800 m/s

☯︎ Vᴇʟᴏᴄɪᴛʏ ᴏғ ʙᴜʟʟᴇᴛ ᴀғᴛᴇʀ ᴘᴀssɪɴɢ ᴛʜʀᴏᴜɢʜ ᴛʜᴇ ᴍᴜᴅ ᴡᴀʟʟ (ᴠ) = 100 m/s

☯︎ Dɪsᴛᴀɴᴄᴇ ᴄᴏᴠᴇʀᴇᴅ ʙʏ ʙʏ ᴛʜᴇ ʙᴜʟʟᴇᴛ (s) = 100ᴄm = 1m

ʟᴇᴛ,

• ᴀᴠᴇʀᴀɢᴇ ʀᴇsɪsᴛᴀɴᴄᴇ ᴏғғᴇʀᴇᴅ ʙʏ ᴛʜᴇ ᴡᴀʟʟ = $\bf\red{F}$

✯ Aᴄᴄᴏʀᴅɪɴɢ ᴛᴏ ᴡᴏʀᴋ-ᴇɴᴇʀɢʏ ᴛʜᴇᴏʀᴇᴍ,

$\begin{cases} \text{Work done by} \\ \text{resistance offered} \\ \text{by mud wall} \end{cases}\Bigg\} = \begin{cases} \text{Decrease in} \\ \text{kinetic energy}\end{cases}\Bigg\}$

$\huge\red\checkmark$ $\bf\purple{F\:.\:s\:=\:\dfrac{1}{2}\:m\:(u^2\:-\:v^2)\:}$

✯ Nᴏᴡ, ᴀᴠᴇʀᴀɢᴇ ʀᴇsɪsᴛᴀɴᴄᴇ ɪs

$\bf\pink{:\implies\:F\:=\:\dfrac{m\:(u^2\:-\:v^2)}{2\:s}\:}$

$\rm{:\implies\:F\:=\:\dfrac{0.02\:\times\:\Big\{(800)^2\:-\:(100)^2\Big\}}{2\:1}\:}$

$\rm{:\implies\:F\:=\:\dfrac{0.02\times{(640000\:-\:10000)}}{2}\:}$

$\rm{:\implies\:F\:=\:\dfrac{0.02\times{630000}}{2}\:}$

$\rm{:\implies\:F\:=\:\dfrac{12600}{2}\:}$

$\bf\green{:\implies\:F\:=\:6300\:N\:}$

$\huge\red\therefore$ [B] Tʜᴇ ᴀᴠᴇʀᴀɢᴇ ʀᴇsɪsᴛᴀɴᴄᴇ ᴏғ ᴛʜᴇ ᴡᴀʟʟ ɪs "6300 N" .