Answer:
The magnitude of the impulse experienced by the particle is 100 kg.m/s.
Explanation:
Given;
mass of the particle, m = 5 kg
initial velocity of the particle, v₁ = 10 m/s
assuming the particle rebounds with same velocity backwards, v₂ = - 10 m/s
The impulse experienced by the particle is the change in linear momentum;
J = ΔP = mv₁ - mv₂
J = m(v₁ - v₂)
J = 5 (10 - (-10))
J = 5 (10 + 10)
J = 5(20)
J = 100 kg.m/s
Therefore, the magnitude of the impulse experienced by the particle is 100 kg.m/s.
