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
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/s².
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.
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
➠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)
Force, F = ??
F = ma
(or) F = m × (v² - u²/2S)
v² - u² = 2aS
a = v² - u²/2S
F = 4 × 10⁻³ × [(0)² - (50)²/2 × 10 × 10²]
F = 4 × 10⁻³ [- 2500 × 10²/2 × 10]