A bullet of 25 gm with a speed 200 m/s stops at a distance of 5 cm. Find the resistive force
Answers
Answer:
The average resistance force is 10000 N.
Explanation:
Given : A bullet of mass 25g moving with a speed 200 m/s is stopped within 5 cm of the target.
To find : What is average resistance force offered by the target?
Solution :
The mass of the bullet is 25 g.
The initial velocity of the bullet is u=200 m/s.
The final velocity is v=0 m/s.
The distance covered is 5 cm.
Into m, 5 cm=0.05 m.
First we find the acceleration.
Using formula,
2as=v^2-u^22as=v
2
−u
2
Substitute the values,
2a(0.05)=0^2-200^22a(0.05)=0
2
−200
2
0.1a=-400000.1a=−40000
a=-\frac{40000}{0.1}a=−
0.1
40000
a=-400000m/s^2a=−400000m/s
2
Converting mass into kg,
25 g= 0.025 kg.
Now, The average resistance force offered by the target is
F=m\times aF=m×a
F=0.025\times 400000F=0.025×400000
F=10000F=10000
Therefore, The average resistance force is 10000 N.