a 120kg ball moving at 18m/s strikes a wall perpendicularly and rebounds straight back at 12m/s. after the initial contact, the centre of the ball moves 0.27cm closer to the wall. Assuming uniform deceleration, show that the time of the contact is 0.00075s. How large an average forced does the ball exert on the wall?
what threw me off is the the time, i can get the average force no problem after i have the time.
my question is do i put the time (0.00075) into the equation s=ut+1/2at(squared), because i have done that which worked out wrong.
or
do i rearrange and substitute some other formulas? if i don't have any formulas needed please tell me!
Vf^2 = vi^1 + 2ad solve for a
but a= deltav/t
Now you can solve for time...
To find the time of contact and the average force, we can use equations of motion. Let's break it down step by step.
Step 1: Find the acceleration.
Since the ball rebounds and moves straight back, the change in velocity is simply the initial velocity minus the final velocity: Δv = 18 m/s - (-12 m/s) = 30 m/s.
Using the equation of motion v = u + at (where v is final velocity, u is initial velocity, a is acceleration, and t is time), we can rearrange it to solve for acceleration:
a = (v - u) / t.
Substituting the values, we get a = (30 m/s) / t.
Step 2: Find the distance traveled during the contact.
The center of the ball moves 0.27 cm closer to the wall during the contact. Convert this distance to meters: 0.27 cm = 0.0027 m.
Using the equation of motion s = ut + (1/2)at^2 (where s is distance, u is initial velocity, a is acceleration, and t is time), we can rearrange it to solve for time:
t = (s - ut) / (1/2)at^2.
Since we are assuming uniform deceleration, the initial velocity (u) is simply the magnitude of the initial velocity: |u| = 18 m/s.
Substituting the known values, we get:
0.0027 m = 18 m/s * t - (1/2)(30 m/s) * t^2.
Rearrange the equation to form a quadratic equation: (1/2)(30 m/s) * t^2 - 18 m/s * t + 0.0027 m = 0.
Step 3: Solve the quadratic equation to find the time.
To solve the quadratic equation, you can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a), where a = (1/2)(30 m/s), b = -18 m/s, and c = 0.0027 m.
Plugging in the values, you get:
t = [18 m/s ± √((18 m/s)^2 - 4 * (1/2)(30 m/s) * 0.0027 m)] / [2 * (1/2)(30 m/s)].
Now simplify the equation and solve for t. Note that we will only consider the positive root since time cannot be negative.
t = 0.00075 s (rounded to 5 decimal places).
Therefore, the time of contact is approximately 0.00075 seconds.
Step 4: Calculate the average force exerted on the wall.
To calculate the average force exerted on the wall, we can use Newton's second law of motion: F = ma (where F is force, m is mass, and a is acceleration).
Since we know the ball's mass is 120 kg and we calculated the acceleration in Step 1, we can substitute those values to find the average force:
F = (120 kg) * (30 m/s) / 0.00075 s.
Calculating the expression gives us:
F ≈ 4,800,000 N.
Therefore, the average force exerted by the ball on the wall is approximately 4.8 million Newtons.