Posted by Audrey on .
A rubber ball (mass 0.30 kg) is dropped from a height of 2.0 m onto the floor. Just after bouncing from the floor, the ball has a speed of 3.9 m/s.
(a) What is the magnitude and direction of the impulse imparted by the floor to the ball?
MAGNITUDE: 3.05 kg m/s
V1 (velocity before) = the square root of 2gh,
v1=square root 2 times gravity(9.8) times the height given, in this case, 2.0m.
v1= sq root (2)(9.8m/s)(2m) =6.260990337
NEXT: use the ans you just got, and plug it in to this equation, which would give you the magnitude.
v1m+v2m=
The ans for v1, or the velocity before the collision, multiplied by the mass, added to the velocity after the collision which is mulitipled by the mass.
(6.2609m/s)*(.30kg)+(3.9 m/s)*(.30 kg)
if you do it sets, v1*m, (6.2609 m/s)*(.30kg) =1.878....
v2*m (3.9 m/s) * (.30 kg) = 1.17 kg m/s
finally, put the two sets together, if you did it seperatly,
1.878 + 1.17= 3.048 kg m/s
rounded, 3.05 for P
DIRECTION : UPWARD
(b) If the average force of the floor on the ball is 16 N, how long is the ball in contact with the floor?
if
P
 = N
T
divide p on both sides, to solve for T, time, using simple algebra.
1 N
 = 
T P
Next, subsitute the value of the variable, STVOTV.
You want to fill in what we know, we know N= 16, and P=3.05, (what we just solved for), and we know the number 1 is there.
With the info we have,
1 16
 = 
T 3.05
16 divided by 3.05=
16/3.05 = 5.245....
LAST STEP, notice
1

T
divide 1 by your answer, (5.245...)=
1 / 5.245 = .190625 s
the
answer is .190625 s
rounded, .01906