When the lever is pulled, 2 kg of carbon dioxide is ejected at a speed of 60 m/s. the remaining mass of the person, chair, and cylinder is 80 kg.

after the ejection, how fast will the chair be moving?

select the best answer from the choices provided.

a)2.0 m/s

b)0.5 m/s

c)1.0 m/s

d)1.5 m/s

d)1.5 m/s

Explanation:

Momentum: Momentum can be defined as the product of mass and velocity.

The S.I unit of momentum is kgm/s. Momentum is a vector quantity.

Momentum of the carbon dioxide = momentum of the chair.

mv = MU Equation 1

making U the subject of the equation above,

U = mv/M Equation 2

where m = mass of the carbon-dioxide, v = velocity of the carbon-dioxide, M = mass of the chair, U = velocity of the chair

Given: m = 2 kg, v = 60 m/s, M = 80 kg

Substituting these values into equation 2

U = 2×60/80

U = 120/80

U = 1.5 m/s

Therefore the speed of the chair after ejection = 1.5 m/s

The right option is d)1.5 m/s