Question

An object with a mass of `4 kg` is traveling at `1 m/s`. If the object is accelerated by a force of `f(x) = x^2 -x +1` over `x in [1, 9]`, where x is in meters, what is the impulse at `x = 2`?

Answer 1

Impulse `J = \Delta p = 1.5377 N.s`

Explanation:

Impulse-Momentum Theorem: `\vec{J_{}}=\Delta\vec{p_{}}`,

To calculate the Impulse ( `\vec{J}`) just calculate the change in momentum ( `\Delta\vec{p_{}}` ). To calculate the change in momentum, remember the relation between momentum (`p`) and the kinetic energy (`K`) :
`K=p^2/(2m) \rightarrow p=\sqrt(2mK)`
Therefore `\Delta p = \sqrt(2m)(\sqrt(K_f)-\sqrt(K_i))`

Work-Energy Theorem: The total work done by all the forces acting on the object must be equal to the change in its kinetic energy `W=\Delta K = K_f-K_i`, where `K_i` and `K_f` are the initial and final kinetic energies.
`K_f = K_i + W`

`K_i = 1/2 mv_i^2 = 1/2(4kg)(1m/s)^2=2` Joules,
`W=\int_1^2F(x)dx=\int_1^2(x^2-x+1)dx=[x^3/3-x^2/2+x]_1^2`
`\qquad\quad` `=(2^3/3-2^2/2+2)-(1/3-1/2+1)=11/6` Joules
`K_f = K_i + W = 2+11/6=23/6` Joules

Now we are ready to evaluate the impulse:
`J=\Delta p = \sqrt(2m)(\sqrt(K_f)-\sqrt(K_i)) = 1.5377 N.s`