# The kinetic energy of an object with a mass of 2 kg constantly changes from 32 J to 136 J over 4 s. What is the impulse on the object at 1 s?

Apr 13, 2016

I am facing this type of question but no one has answered it yet.
Probably the question is not clear enough as impulse at a particular time is wanted here , which is not possible by definition

Now I place here a probable solution considering that the impulse at a particular time (say Tsec) means

Either (1) It is the change in momentum during $t = 0 \to t = T \sec$ (this is by definition )

Or (2) It is the momentum of the body at $T s$ ( assumed )

In our question the KE of an object of mass $m = 2 k g$ constantly changes from $32 J \to 135 J$ in $4 s$. So during this time KE vs time graph will be straight line having a + slope
So slope $= \frac{\Delta E}{\Delta t} = \frac{{E}_{f} - {E}_{i}}{t - 0} = \frac{136 - 32}{4 - 0} = 26$

If the velocity of the object is v at $t$ th sec Then
we can write
$\frac{{E}_{t} - {E}_{i}}{t - 0} = 26$
$\implies \frac{\frac{1}{2} \times 2 {v}^{2} - 32}{t - 0} = 26$
$\implies {v}^{2} = 26 t + 32$
$\implies v = \sqrt{26 t + 32}$
So the momentum at $t = 0 s$
${p}_{0} = 2 \times {v}_{0} = 2 \times \sqrt{32} k g \frac{m}{s} =$
So the momentum at $t = 1 s$
${p}_{1} = 2 \times {v}_{1} = 2 \times \sqrt{26 \times 1 + 32} = 2 \sqrt{58} k g \frac{m}{s}$

Hence impulse during 1s = ${p}_{1} - {p}_{0} = 2 \left(\sqrt{58} - \sqrt{32}\right) k g \frac{m}{s}$

OR otherwise impulse at 1s is momentum at is $= {p}_{1} = 2 \sqrt{58} k g \frac{m}{s}$