A bullet of mass 10 g hits a target and penetrated 2 cm in to it if the average resistance offered by target is 100 N then the velocity with which the bullet hits the target is ?

100 N * .02 m = 2 Joules = work done

so

(1/2)(0.010) v^2 = 2

v^2 = 400

v = 20 meters/second

Right answer

To find the velocity with which the bullet hits the target, we can use the principle of work and energy.

The work done on the bullet by the resistance force offered by the target is equal to the change in kinetic energy of the bullet.

The work done by a force can be calculated using the formula:

Work = Force × Displacement × cos(θ)

In this case, the force is the resistance force offered by the target (100 N) and the displacement is the penetration depth of the bullet into the target (2 cm or 0.02 m). The angle between the force and displacement vectors is assumed to be zero, so we can omit the cos(θ) term.

Now we can calculate the work done on the bullet:

Work = 100 N × 0.02 m = 2 J (Joules)

The work done on the bullet is also equal to the change in kinetic energy of the bullet:

Work = ΔKE = (1/2)mv^2

Here, m is the mass of the bullet (10 g or 0.01 kg) and v is the velocity with which the bullet hits the target.

Rearranging the formula, we can solve for v:

v^2 = (2 × 2 J) / (0.01 kg)
v^2 = 400 m^2/s^2

Taking the square root of both sides, we find:

v = 20 m/s

Therefore, the velocity with which the bullet hits the target is 20 m/s.