7.A bullet of 10 g strikes a sand bag at a speed of 1000m/s and gets embedded after traveling 5cm.(a) Calculate the resistive force exerted by the sand on the bullet.(b) Time taken by the bullet to come to rest. *
Answers
Given:-
→Mass of the bullet = 10g
→Initial velocity of the bullet = 1000m/s
→Distance travelled by the bullet = 5cm
To find:-
→The resistive force exerted by the sand on the bullet.
→Time taken by the bullet to come to rest.
Solution:-
•Final velocity of the bullet will be zero as it finally comes to rest after embedding into the sand bag.
Firstly, let's convert the mass of the bullet from g to kg and distance travelled by the bullet from
cm to m.
=>1g = 0.001kg
=>10g = 10(0.001)
=>0.01kg
=>1cm = 0.01m
=>5cm = 5(0.01)
=>0.05m
By using the 3rd equation of motion,we get:-
=>v²-u² = 2as
=>0-(1000)² = 2×a×0.05
=>a = -1000×1000/2×0.05
=>a = -10⁷m/s²
Hence,we have got acceleration of the bullet as -10⁷m/s².
By using Newton's 2nd law of motion,we get:-
=>F = ma
=>F = 0.01(-10⁷)
=>F = -10⁵N
[Here,negative sign represents
resistive force]
Now,by using the 1st equation of motion,we get:-
=>v = u+at
=>0 = 1000+(-10⁷)t
=>-1000 = -10⁷t
=>t = -1000/-10⁷
=>t = 10⁻⁴s
Thus:-
→Resistive force exerted by sand on the bullet is -10⁵N.
→Time taken by the bullet to come to rest is 10⁻⁴s.
Answer:
(i) the resistive force exerted by the sand on the bullet is - 1 x 10⁵ N
(ii) the time taken by the bullet to come to rest is 1 x 10⁻⁴ s
Given Parameter :
- mass of bullet, m = 10 g = 0.01 kg
- speed of the bullet, v = 1000 m/s = 10³ m/s
- distance traveled by the bullet, d = 5 cm = 0.05 m
Explanation:
(i) the resistive force exerted by the sand on the bullet
the acceleration of the bullet is given by;
→ v² = u² + 2as
→ 0 = (10³)² + a(2 x 0.05)
→ -0.1a = (10⁶)
→ -a = (10⁶) / 0.1
→ a = -1 x 10⁷ m/s²
The resistive force is given by;
➳ F = ma
➳F = (0.01)(-1 x 10⁷)
➳ F = - 1 x 10⁵ N
(ii) the time taken by the bullet to come to rest.
➝ v = u + at
➝ 0 = 10³ + (-1 x 10⁷ t)
➝ 1 x 10⁷ t = 10³
➝ t = 10³ / 10⁷
➝ t = 1 x 10⁻⁴ s