A 70-g ball B dropped from a height h_0=1.5 m reaches a height h_2=0.25 m after bouncing twice from identical 210-g plates. Plate A rests directly on hard ground, while plate C rests on a foam-rubber mat. Determine
(a) the coefficient of restitution between the ball and the plates,
(b) the height h_1 of the ball’s first bounce.
To determine the coefficient of restitution between the ball and the plates, we need to use the equation:
e = √(h2 / h1)
where e is the coefficient of restitution, h2 is the final height after the second bounce, and h1 is the initial height before the first bounce.
Given that the initial height (h0) is 1.5 m and the final height after the second bounce (h2) is 0.25 m, we can calculate h1 using the equation for conservation of mechanical energy:
mgh0 = mgh1 + mgh2
where m is the mass of the ball, g is the acceleration due to gravity (9.8 m/s^2).
Let's calculate h1 first:
mgh0 = mgh1 + mgh2
(70 g) * (9.8 m/s^2) * (1.5 m) = (70 g) * (9.8 m/s^2) * h1 + (70 g) * (9.8 m/s^2) * (0.25 m)
Simplifying the equation:
102.9 = 68.2 h1 + 1.715
Now, let's solve for h1:
102.9 - 1.715 = 68.2 h1
101.185 = 68.2 h1
h1 = 101.185 m / 68.2
h1 ≈ 1.483 m
Now that we have h1 ≈ 1.483 m and h2 = 0.25 m, we can calculate the coefficient of restitution:
e = √(h2 / h1)
e = √(0.25 / 1.483)
e ≈ √(0.1688)
e ≈ 0.411
Therefore, the coefficient of restitution between the ball and the plates is approximately 0.411.
To summarize:
(a) The coefficient of restitution between the ball and the plates is approximately 0.411.
(b) The height of the ball's first bounce, h1, is approximately 1.483 m.
To determine the coefficient of restitution between the ball and the plates and the height of the ball's first bounce, we can use the principles of conservation of energy and momentum.
(a) Calculating the coefficient of restitution:
The coefficient of restitution (e) is a measure of the elasticity of a collision. It is defined as the ratio of the relative velocity of separation to the relative velocity of approach between two objects after a collision.
In this case, the ball bounces twice from identical plates. Since the plates are identical, the coefficient of restitution between the ball and each plate is the same. Let's denote the coefficient of restitution as 'e'.
After the first bounce, the ball loses some energy due to the collision and the bouncing process. The loss of energy is given by (1 - e^2) where e is the coefficient of restitution.
Since the ball reaches a height of h_2 = 0.25 m after the second bounce, we can calculate the energy loss as follows:
Energy loss = (1 - e^2) * (Potential energy at h_0 - Potential energy at h_2)
= (1 - e^2) * (m * g * (h_0 - h_2))
Substituting the known values:
Energy loss = (1 - e^2) * (0.07 kg * 9.8 m/s^2 * (1.5 m - 0.25 m))
= (1 - e^2) * (0.07 kg * 9.8 m/s^2 * 1.25 m)
Now, we know that the energy loss is equal to the initial kinetic energy of the ball after being dropped from a height of h_0. The initial kinetic energy is given by (1/2) * m * v^2, where 'v' is the initial velocity of the ball just before it hits the plate A.
Energy loss = (1/2) * m * v^2
We substitute the known values and solve for 'v':
(1 - e^2) * (0.07 kg * 9.8 m/s^2 * 1.25 m) = (1/2) * 0.07 kg * v^2
Solving for 'v':
v^2 = [2 * (1 - e^2) * (0.07 kg * 9.8 m/s^2 * 1.25 m)] / 0.07 kg
Simplifying,
v^2 = 2 * (1 - e^2) * 9.8 m/s^2 * 1.25 m
v^2 = 24.5(1 - e^2)
Now, we can calculate the velocity just before the ball hits the plate C. The energy loss during the second bounce is similar to the energy loss during the first bounce. Therefore, we can write:
v^2 = [2 * (1 - e^2) * (0.07 kg * 9.8 m/s^2 * (h_0 - h_1))] / 0.07 kg
Simplifying,
v^2 = 2 * (1 - e^2) * 9.8 m/s^2 * (h_0 - h_1)
Now, we have two equations for v^2, so we can equate them and solve for 'e':
24.5(1 - e^2) = 9.8 * (h_0 - h_1)
Substituting the known values,
24.5(1 - e^2) = 9.8 * (1.5 - h_1)
Now, solve for 'e'.
(b) Calculating the height of the ball's first bounce (h_1):
Once we have the value of the coefficient of restitution 'e', we can substitute it into the equation we derived for the velocity just before the ball hits plate C:
v^2 = 2 * (1 - e^2) * 9.8 m/s^2 * (h_0 - h_1)
Substituting the known values and the calculated value of 'e', we can solve for h_1:
v^2 = 2 * (1 - e^2) * 9.8 m/s^2 * (1.5 m - h_1)
Simplifying further,
v^2 = 29.4 * (1 - e^2) * (1.5 m - h_1)
Now, we can solve for 'h_1'.
By substituting the calculated value of 'e' into the equation, we can determine the coefficient of restitution between the ball and the plates. Similarly, by solving the equation for 'h_1', we can determine the height of the ball's first bounce.
I do not care what the mass is, call it m
If it bounces off the same thing twice (I do not get it about the two plates because you provide data for only one of them)
v = velocity at impact with plate A
(1/2) m v^2 = m g h
v = sqrt(2 gh) = sqrt (2*9.81*1.5) = sqrt (29.43) = 5.42 m/s
w = velocity headed up from first hit = 5.42 k
where k is our coef of restitution
we lose no energy going up and then coming down so
we hit plate B at speed w = 5.42 k
we rebound from plate B at speed x = 5.42 k^2
now
m g h = m g (0.25) = (1/2) m (5.42k^2)^2
9.81 (.25) = .5 (29.4) k^4
k^4 = .167
k^2 = .408
k = .639
That should get you started although I am not sure what the two plates is about.