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?

Apr 30, 2016

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

$J = 5.57$ $k g \cdot \frac{m}{s}$

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:

$46.1 \cdot 3 = 138.3$ $J$

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

${K}_{3 s} = {K}_{i} + {K}_{c h} = 243 + 138.3 = 381.3$ $J$

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

$K = \frac{1}{2} \cdot m \cdot {u}^{2}$

$u = \sqrt{\frac{2 K}{m}}$

${u}_{1} = \sqrt{\frac{2 \cdot 243}{1}} = 22.05 \frac{m}{s}$

${u}_{2} = \sqrt{\frac{2 \cdot 381.3}{1}} = 27.62 \frac{m}{s}$

Finally, the impulse can be calculated:

J=Δp=m*u_2-m*u_1=1*27.62-1*22.05

$J = 5.57$ $k g \cdot \frac{m}{s}$