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A crane drops a 0.3 kg steel ball onto a steel plate. The ball's speeds just before impact and after are 4.5 m/s and 4.2 m/s, respectively.If the ball is in contact with the plate for 0.03 s, what is the magnitude of the average force that the ball exerts on the plate during impact.

Equations I thought would be useful (Basic Physics):

F=ma

a=v/t

My ATTEMPT at the question:

Well first i took my givens which were:

Mass of Ball: 0.3 kg

Velocity Before impact: 4.5 m/s

Velocity After impact: 4.2 m/s

Accelration of gravity constant: 9.8 m/s^2 (^2=squared[Exponential value as it is written on a calculator)

So my thinking is that we should find the easiest force equation there, being the first one (before impact):

F1=ma

F1=(0.3)x(9.8)

F1=2.94N (Newtons)

Then the second one (after impact):

F2=ma

F2=(0.3)a

(Finding acceleration)a=v/t

a=(4.2-4.5)/(0.03)

a=-3m/s^2

(Back to Force of second equation)

F2=(0.3)x(-3)

F2=-3N(Newtons)

So now i assume that force is a scalar quantity and thus negatives don't really apply, only the value so i use:

ƩF=Average Force=(F1+F2)/2

ƩF=(3+2.94N)/2

ƩF=(5.94N)/2

ƩF=2.97N and the multiple choice answers are:

(a) 3.0N (MY ANSWER SINCE I ROUNDED 2.97N TO 3.0N)

(b) 87N

(c) 3.5N

(d) 30N

(e) 133N

So I ask is how my answer (which was wrong) is not correct, where i messed up, what formulas i should have used, and what the correct answer is. (Note: This isn't homework it is a quiz question that i took and i never understood how it was wrong.)