A 5kg particle moving at a speed of 10m/s to the right makes an elastic collision with a wall and rebounds backward calculate the magnitude of the impulse of the body

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.

the total distance traveled by the car is

30km + 49km = 79 km .

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s