# The kinetic energy of an object with a mass of 2 kg constantly changes from 8 J to 136 J over 4 s. What is the impulse on the object at 1 s?

##### 1 Answer
Aug 20, 2016

${\vec{J}}_{0 \to 1} = 4 \left(\sqrt{10} - \sqrt{2}\right) \hat{p}$ N s

#### Explanation:

I think there is something wrong in the formulation of this question.

With Impulse defined as
$\vec{J} = {\int}_{t = a}^{b} \vec{F} \left(t\right) \setminus \mathrm{dt}$
$= {\int}_{t = a}^{b} \vec{\dot{p}} \left(t\right) \setminus \mathrm{dt} = \vec{p} \left(b\right) - \vec{p} \left(a\right)$

then the Impulse on the object at t= 1 is

$\vec{J} = {\int}_{t = 1}^{1} \vec{F} \left(t\right) \setminus \mathrm{dt} = \vec{p} \left(1\right) - \vec{p} \left(1\right) = 0$

It may be that you want the total impulse applied for $t \in \left[0 , 1\right]$ which is

$\vec{J} = {\int}_{t = 0}^{1} \vec{F} \left(t\right) \setminus \mathrm{dt} = \vec{p} \left(1\right) - \vec{p} \left(0\right) q \quad \star$

To evaluate $\star$ that we note that if the rate of change of kinetic energy $T$ is constant, ie:

$\frac{\mathrm{dT}}{\mathrm{dt}} = c o n s t$

then

$T = \alpha t + \beta$

$T \left(0\right) = 8 \implies \beta = 8$

$T \left(4\right) = 136 = \alpha \left(4\right) + 8 \implies \alpha = 32$

$T = 32 t + 8$

Now $T = {\left\mid \vec{p} \right\mid}^{2} / \left(2 m\right)$.

$\implies \left(\vec{p} \cdot \vec{p}\right) = 4 \left(32 t + 8\right)$

$\vec{p} = 2 \sqrt{\left(32 t + 8\right)} \hat{p}$

and
$\vec{p} \left(1\right) - \vec{p} \left(0\right)$

$= \left(2 \sqrt{\left(32 + 8\right)} - 2 \sqrt{8}\right) \hat{p}$

$= 4 \left(\sqrt{10} - \sqrt{2}\right) \hat{p}$ N s