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.

report!

explanation:

physics means: "the branch of science concerned with the nature and properties of matter and energy. the subject matter of physics, distinguished from that of chemistry and biology, includes mechanics, heat, light and other radiation, sound, electricity, magnetism, and the structure of atoms."

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form