Question

The kinetic energy of an object with a mass of `1 kg` constantly changes from `243 J` to `658 J` over `9 s`. What is the impulse on the object at `3 s`?

Answer 1

You must aknowledge that the key words are "constantly changes". Afterwards, use the kinetic energy and impulse definitions.

Answer is:

`J=5.57` `kg*m/s`

Explanation:

The impulse is equal to the change of momentum:

`J=Δp=m*u_2-m*u_1`

However, we are missing the velocities.

Constantly changing means that it changes "steadily". This way, we can assume that the rate of change of the kinetic energy `K` with respect to time is constant:

`(ΔK)/(Δt)=(658-243)/9=46.1 J/s`

So for every second the object gains `46.1` joules. For three seconds:

`46.1*3=138.3` `J`

Therefore the kinetic energy at `3s` is equal to the initial plus the change:

`K_(3s)=K_(i)+K_(ch)=243+138.3=381.3` `J`

Now that both kinetic energies are known, their velocities can be found:

`K=1/2*m*u^2`

`u=sqrt((2K)/m)`

`u_1=sqrt((2*243)/1)=22.05m/s`

`u_2=sqrt((2*381.3)/1)=27.62m/s`

Finally, the impulse can be calculated:

`J=Δp=m*u_2-m*u_1=1*27.62-1*22.05`

`J=5.57` `kg*m/s`