Answer:
(a)

(b)
1120 N
Explanation:
Change in velocity,
is given by subtracting the initial velocity from the final velocity and expressed as 
Where v represent the velocity and subscripts f and i represent final and initial respectively. Since the ball finally comes to rest, its final velocity is zero. Substituting 0 for final velocity and the given figure of 8 m/s for initial velocity then the change in velocity is given by

To find
then we substitute 7 kg for m and -8 m/s for
therefore 
(b)
The impact force, F is given as the product of mass and acceleration. Here, acceleration is given by dividing the change in velocity by time ie

Substituting t with 0.05 s then 
Since F=ma then substituting m with 7 Kg we get that F=7*-160=-1120 N
Therefore, the impact force is equivalent to 1120 N
Answer:
The first law, also called the law of inertia, was pioneered by Galileo. This was quite a conceptual leap because it was not possible in Galileo's time to observe a moving object without at least some frictional forces dragging against the motion. In fact, for over a thousand years before Galileo, educated individuals believed Aristotle's formulation that, wherever there is motion, there is an external force producing that motion.
The second law, $ f(t)=m\,a(t)$ , actually implies the first law, since when $ f(t)=0$ (no applied force), the acceleration $ a(t)$ is zero, implying a constant velocity $ v(t)$ . (The velocity is simply the integral with respect to time of $ a(t)={\dot v}(t)$ .)
Newton's third law implies conservation of momentum [138]. It can also be seen as following from the second law: When one object ``pushes'' a second object at some (massless) point of contact using an applied force, there must be an equal and opposite force from the second object that cancels the applied force. Otherwise, there would be a nonzero net force on a massless point which, by the second law, would accelerate the point of contact by an infinite amount.
Explanation:
Acceleration of cheetah (a) = 4m/s²
time = 10s
initial velocity(u) = 0
final velocity = v
distance travelled = s
v = u +at = 0 + 10×4 = 40m/s
s = (v²-u²)/2a = 40²/(2×4) = 1600/8 = 200m