Ascientist launches a 100 g ball from a catapult as part of an experiment. the ball travels at 500 m/s after launch. the catapult weighs 2,000 g and has a slight recoil. which relationship best describes the recoil of the catapult

V₂ = - m₁ V₁ / m₂

Explanation:

According to law of conservation of momentum, "Total momentum of an isolated system remains constant. i.e

Pi = Pf

We consider ball and catapult an isolated system.

before launching ball momentum of the system is zero.

After launching ball momentum of ball is:

Pb= 0.1 * 500 = 50 kg m/s

Now according to law of conservation of momentum:

Pf = Pi

⇒ Pb + Pc = 0

Let Pb= m₁ V₁

&    Pc = m₂ V₂

So

m₁ V₁ + m₂ V₂ = 0

⇒ V₂ = - m₁ V₁ / m₂

The negative sing shows that catapult velocity will have opposite direction to the ball velocity.

a.) travels faster in solids because the particles are closer together.

air is a gas water is a liquid

answer; equivalent to a one(1) horsepower motor;