A bullet of mass 15g travelling horizontally with a velocity of 150 m/s strikes a stationary wooden block and comes to rest in 0.03 seconds. The distance of penetration of the bullet into the block will be: Options 0 2.60m o 7.75 m O 5.0 m O 2.25m
Answers
Answer :-
2.25 m
Given :-
- Mass of bullet is m= 15g
- Initial velocity is u=150m/s
- It comes to rest i.e Final velocity (v=0)
- Time taken is t= 0.03 s
To find :-
Distance travelled .
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Solution:-
Firstly let's find the acceleration of the bullet.
a = v-u/t
where,
- a = acceleration
- v = final velocity
- u= initial velocity
- t= time taken
⇒a= (0 - 150)/ 0.03
⇒a = -150/3× 10^-2
⇒a = -50 × 10²
⇒a= -5 × 10 × 10²
⇒a = -5× 10³
⇒a = -5× 1000
⇒a = -5000 m/s²
By using third equation of motion
v² - u² = 2as
where,
- v = final velocity
- u = initial velocity
- a = acceleration
- s = distance covered
⇒(0)² - (150)² = 2(-5000) s
⇒0 - 22500 = -10,000 s
⇒-22,500 = -10,000 s
⇒225 = 100 s
⇒s = 225/100
s = 2.25m
So, the distance covered is 2.25 m
The correct answer to the above question is (d), which is 2.25m, and the explanation is given below -
Given: mass of bullet = 15 grams
The velocity of bullet = 150m/s
Final velocity = 0
Time = 0.03 seconds
To find: Distance traveled
Solution:
First, we will find the acceleration of the bullet,
Initial velocity = 150 m/s, Final velocity = 0
Time taken = 0.03 seconds
Acceleration produced by the bullet will be equal to rate of change of velocity.
a = (final velocity - initial velocity)/time taken
a = (0 - 150)/ 0.03
= - 5000 m/s²
The retardation produced by the bullet will be 5000 m/s²
Now, by the law of kinetics,
v² - u² = 2as
Distance of penetration of the bullet into the block will be
0 - (150)² = 2×5000s
s = 2.25 m
Therefore, the distance traveled inside the block will be 2.25m