A bullet of 0.02 kg moving with a speed of 100 ~ms ⁻¹ hits a surface and comes to rest in 0.1 s What is the magnitude of the impulse of the force?

Step-by-step solution

The impulse J is defined as the change in linear momentum of the object. J = p = m(v_f - v_i) Given mass m = 0.02 kg , initial velocity v_i = 100 ms ⁻¹ , and final velocity v_f = 0 ms ⁻¹ (since the bullet comes to rest). J = 0.02 (0 - 100) J = 0.02 (-100) = -2 kg ms ⁻¹ The magnitude of the impulse is |J| = |-2| = 2 N.s . Answer: 2 N.s