Question

and bullet of mass 5 grams travels with a speed of 500 metre per second if it penetrates a fixed target which offers constant resistive force of 2000 Newton to the motion of the bullet find the initial kinetic energy of the bullet and also find the distance to which the bullet has penetrated​

Answer 1

Answer:formulA= m1u2+m2u2=m1v1+m2v2

Explanation:

Solve this question with this formula

Answer 2

Answer:

1)625J

2)0.3125m

Explanation:

1st convert all the units to si system

so 5g=0.005kg

the initial kinetic energy is

1/2 ×mv^2

Therefore it is 1/2 × 0.005×250000

simplifying this will be

625J

now we know that change in kinetic energy is 625 (final KE=0)

therefore, work done by the target will be 2000×s

so we can equate 625=2000×s

s=0.3125m