Answer:
Approximately [tex](-4.9\times 10^{2}\; {\rm N})[/tex] (rounded to two significant figures.)
Explanation:
The impulse [tex]\mathbf{J}[/tex] on an object is equal to the change in the momentum [tex]\mathbf{p}[/tex] of this object. Dividing impulse by duration [tex]\Delta t[/tex] of the contact would give the average external force that was exerted on this object.
If an object of mass [tex]m[/tex] is moving at a velocity of [tex]\mathbf{v}[/tex], the momentum of that object would be [tex]\mathbf{p} = m\, \mathbf{v}[/tex].
Let [tex]m[/tex] denote the speed of this ball. Let [tex]\mathbf{v}_{0}[/tex] denote the velocity of this ball before the contact, and let [tex]\mathbf{v}_{1}[/tex] denote the velocity of the ball after the contact.
Momentum [tex]\mathbf{p}_{0}[/tex] of this ball before the contact: [tex]\mathbf{p}_{0} = m\, \mathbf{v}_{0}[/tex].
Momentum [tex]\mathbf{p}_{1}[/tex] of this ball after the contact: [tex]\mathbf{p}_{1} = m\, \mathbf{v}_{1}[/tex].
Impulse, which is equal to the change in momentum:
[tex]\mathbf{J} = \mathbf{p}_{1} - \mathbf{p}_{0} = m\, \mathbf{v}_{1} - m\, \mathbf{v}_{0}[/tex].
Since the velocity of the ball after the contact is in the [tex](-x)[/tex] direction, [tex]\mathbf{v}_{1} = -20.0\; {\rm m\cdot s^{-1}}[/tex]. The average external force on this ball would be:
[tex]\begin{aligned} \mathbf{F} &= \frac{\mathbf{J}}{\Delta t} \\ &= \frac{m\, \mathbf{v}_{1} - m\, \mathbf{v}_{0}}{\Delta t} \\ &= \frac{m\, (\mathbf{v}_{1} - \mathbf{v}_{0})}{\Delta t} \\ &= \frac{0.145\; {\rm kg} \times ((-20.0\; {\rm m\cdot s^{-1}}) - 14\; {\rm m\cdot s^{-1}})}{0.010\; {\rm s}} \\ &\approx -4.9 \times 10^{2}\; {\rm N}\end{aligned}[/tex].
(Rounded to two significant figures.)
