A 10gm bullet moving at 100 m/s strikes a wooden block kept fixed. The bullet comes to rest and stops in 10 cm . What is the average force (in N) experienced by the bullet..??
Answers
Required Answer :
Average force experienced by the bullet = - 500 N
Given :
Mass of the bullet = 10 gram
Initial velocity of the bullet = 100 m/s
Final velocity of the bullet = 0 m/s
Distance = 10 cm
To find :
Average force experienced by the bullet
Solution :
As we know,
- 1 g = 1/1000 kg
In order to convert the value from g into kg, multiply it by 1/1000.
→ Mass = 10 × 1/1000
→ Mass = 0.01 kg
As we know,
- 1 cm = 1/100 m
In order to convert the value from cm into m, multiply it by 1/100.
→ Distance = 10 × 1/100
→ Distance = 0.1 m
Using the third equation of motion :-
- v² - u² = 2as
where,
v = Final velocity
u = Initial velocity
a = Acceleration
s = Distance/Displacement
Substituting the given values :-
→ (0)² - (100)² = 2(a)(0.1)
→ - 10000 = 0.2 × a
→ - 10000 × 10/2 = a
→ - 10000 × 5 = a
→ - 50000 = a
Therefore, the acceleration of the bullet = - 50000 m/s²
Now, using the formula :-
- F = ma
where,
F = Force
m = mass
a = acceleration
Substituting the given values :-
→ Force = 0.01 × (- 50000)
→ Force = 1/100 × - 50000
→ Force = - 500
Therefore, the force = - 500 N