Question

(I) (a) If the kinetic energy of a particle is tripled, by what factor has its speed increased? (b) If the speed of a particle is halved, by what factor does its kinetic energy change?

Short answer

(a) The speed increases by \(\sqrt 3 \) times the initial speed.

(b) The kinetic energy becomes \(\frac{1}{4}\) times the initial kinetic energy.

Step-by-step solution

Relationship between speed and kinetic energy

The kinetic energy of a particle is proportional to the square of the speed of the particle.

Formula for kinetic energy

Let m be the mass of the particle and vbetheinitial speed of the particle.

Therefore, the kinetic energy is:

\(K = \frac{1}{2}m{v^2}\)

Calculation for part (a)

The initial kinetic energy of the particle is \(K = \frac{1}{2}m{v^2}\).

Part (a)

If the kinetic energy of the particle is tripled, the kinetic energy of the particle becomes 3K.

Now, let the speed be \(v'\) when the kinetic energy is tripled.

Therefore, you can write:

\(\begin{aligned}\frac{1}{2}m{\left( {v'} \right)^2} = 3K\\\frac{1}{2}m{\left( {v'} \right)^2} = 3 \times \frac{1}{2}m{v^2}\\{\left( {v'} \right)^2} = 3{v^2}\\v' = \sqrt 3 v\end{aligned}\)

Hence, the speed increases by \(\sqrt 3 \) times the initial speed.

Calculation for part (b)

Now, the speed of the particle is halved; so the new speed of the particle is \(v'' = \frac{v}{2}\) .

Then, the kinetic energy in this case becomes:

\(\begin{aligned}K'' = \frac{1}{2}m{\left( {v''} \right)^2}\\ = \frac{1}{2}m{\left( {\frac{v}{2}} \right)^2}\\ = \frac{1}{4} \times \frac{1}{2}m{v^2}\\ = \frac{K}{4}\end{aligned}\)

Hence, the kinetic energy becomes \(\frac{1}{4}\) times the initial kinetic energy.