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
(a) The magnitude of the impulse imparted by the floor to the ball is 3.05 kg m/s. To calculate this, you first need to find the velocity of the ball before the collision using the equation v1 = √(2gh), where g is the acceleration due to gravity (9.8 m/s²) and h is the height (2.0 m). Plugging in the values, you get v1 = √(2*9.8*2) = 6.261 m/s.
Next, use the formula v1m + v2m = P, where v1 is the initial velocity, v2 is the final velocity (3.9 m/s), m is the mass of the ball (0.30 kg), and P is the impulse. Plugging in the values, you get (6.261 m/s)*(0.30 kg) + (3.9 m/s)*(0.30 kg) = 1.878 + 1.17 = 3.048 kg m/s. Rounded to two decimal places, the magnitude of the impulse is 3.05 kg m/s.
The direction of the impulse is upward since the ball bounces off the floor.
(b) If the average force of the floor on the ball is 16 N, you can use the formula P = FΔt, where P is the impulse (3.05 kg m/s), F is the force (16 N), and Δt is the time. Rearranging the formula to solve for Δt, you get Δt = P/F. Plugging in the values, you get Δt = 3.05 kg m/s / 16 N = 0.190625 s. Rounded to five decimal places, the ball is in contact with the floor for approximately 0.19063 seconds.