Physics, asked by Sahuleeja92, 5 months ago

a bullet of mass 4g fired with a velocity of 50 m/s enters a wall up to a depth of 10 cm. Calculate the average resistance offered by the wall ​

Answers

Answered by Anonymous
42

Given :

  • Mass of bullet (m) = 4 g
  • Initial Velocity (u) = 50 m/s
  • Final Velocity (v) = 0 m/s
  • Depth [Distance] (s) = 10 cm

To Find :

  • The average resistance offered by the wall = ?

Solution :

First of all convert the mass of bullet from g to kg :

→ Mass of bullet (m) = 4 g

→ Mass of bullet (m) = 4 ÷ 1000 kg

Mass of bullet (m) = 0.004 kg

  • Hence,the mass of bullet is 0.004 kg.

Now, convert the depth from cm to m :

⇒Depth (s) = 10 m

⇒Depth (s) = 10 ÷ 100 cm

⇒Depth (s) = 1 ÷ 10 cm

Depth (s) = 0.1 m

  • Hence,the depth is 0.1 m.

Now,let's find the acceleration of bullet by using third equation of motion :

➻ v² - u² = 2as

➻ (0)² - (50)² = 2 × a × 0.1

➻ 0 - 2500 = 0.2a

➻ -2500 ÷ 0.2 = a

➻ -12500 = a

➻ a = - 12500 m/s²

  • Hence,the acceleration of the bullet is - 12500 m/.

Finding force i.e resistance offered by the wall :

➺ Force = mass × acceleration

➺ Force = 0.004 × - 12500

➺ Force = - 50 N

  • Hence,the force exerted by the bullet on the wall is -50 N. So, according to newton's third law the force exerted by the wall on the bullet is +50 N.
Answered by Fαírү
235

\large\bold{\underline{\underline{Question:-}}}

A bullet of mass 4g fired with a velocity of 50 m/s enters a wall up to a depth of 10 cm. Calculate the average resistance offered by the wall

\text{\large\underline{\green{Given:-}}}

➠Mass of the bullet, u = 4 g = 4 × 10⁻³ kg

➠Initial velocity, u = 50 m/s

➠Distance, S = 10 cm = 10 × 10⁻² m

➠Final velocity, v = 0 (since bullet comes to rest finally)

\text{\large\underline{\pink{To \: Calculate:-}}}

Force, F = ??

\text{\large\underline{\orange{Formula \: To \: Be \: used:-}}}

F = ma

(or) F = m × (v² - u²/2S)

v² - u² = 2aS

a = v² - u²/2S

\text{\large\underline{\purple{Solution:-}}}

F = 4 × 10⁻³ × [(0)² - (50)²/2 × 10 × 10²]

F = 4 × 10⁻³ [- 2500 × 10²/2 × 10]

F = - 50 N.

➠ Hence, the average resistances offered by the wall - 50 N.


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