mass of bullet be 0.8 kginitial velocity- 600 m/sfinal velocity _ 0 m/stime taken- 0.13 s v = u+ at0 = 600 + a ( 0.13s )a = -600÷ 0.13a= 4615.3 m/s^2F = ma = 0.8 × 4615.3 = 3692.2 NF i = F × T = 3692.2 × 0.13 s = 480 N

Given:
Mass of bullet (m) = 80 gm = 0.08 kg
Initial velocity (u) = 600 m/s
Final velocity (v) = 0 m/s
Time taken (t) = 0.13 sec

Impulse of force:
Impulse is defined as the change in momentum of an object. Mathematically, impulse (J) is given by the product of force (F) and time (t), which can be expressed as:

J = F * t

The momentum of an object is given by the product of its mass and velocity:

Momentum (p) = m * v

In this case, the initial momentum (p1) of the bullet is:

p1 = m * u

The final momentum (p2) of the bullet is:

p2 = m * v

Since the bullet is brought to rest, the change in momentum (Δp) is given by:

Δp = p2 - p1

Substituting the values:

Δp = m * v - m * u
= m * (v - u)
= 0.08 kg * (0 m/s - 600 m/s)
= -48 kg·m/s

The impulse of force (J) is equal to the change in momentum (Δp):

J = -48 kg·m/s

Average force exerted:
The average force exerted on an object can be calculated by dividing the impulse of force (J) by the time taken (t):

Average force (F_avg) = J / t

Substituting the values:

F_avg = -48 kg·m/s / 0.13 s
= -369.23 N

The negative sign indicates that the force is exerted in the opposite direction of the initial velocity.

Therefore, the impulse of force is -48 kg·m/s and the average force exerted is -369.23 N.