A bullet of mass 50g travelling with a speed of 15m.s penetrates into a bag of sand and is uniformly brought to rest in 0.05 sec. find (a)How for the bullet Will penetrate into the bag of sand?(b) The average force exerted by the sand.
Answers
Answer:
Mass, m=0.05 kg
Initial velocity, u=500 m/s
Distance covered, S=100cm=1 m
Final velocity v=0
Now, from the equation of motion
v
2
−u
2
=2as
0−(500)
2
=2×a×1
a=−125000m/s
2
Now, the average force exerted on the block
F=ma
F=0.05×125000
F=6.25×10
3
F=0.625×10
4
N
Hence, the average force is 0.625×10
4
N
Given :
Mass of bullet = 50g = 0.05kg
Initial velocity = 15m/s
Final velocity = zero
Time interval = 0.05s
To Find :
Distance covered by bullet after penetrating into the bad of sand and average force exerted by the sand.
Solution :
❖ In order to find distance covered by bullet, first we need to find acceleration produced in the bullet. Acceleration can be calculated by using first equation of kinematics
➛ v = u + at
- v denotes final velocity
- u denotes initial velocity
- a denotes acceleration
- t denotes time
By substituting the given values;
➛ 0 = 15 + a(0.05)
➛ 0.05a = -15
➛ a = -15/0.05
➙ a = -300 m/s²
Negative sign shows retardation!
❒ Distance covered by bullet :
➙ v² - u² = 2as
➙ 0² - 15² = 2(-300)s
➙ -225 = -600s
➙ s = 225/600
➙ s = 0.375 m
❒ Average force exerted by the sand :
As per newton's second law of motion, force is measured as the product of mass and acceleration.
Mathematically, F = ma
➨ F = m × a
➨ F = 0.05 × (-300)
➨ F = -15 N (Retarding force)
Average force |F| = 15N