Answer:
398 m/s
Explanation:
The acceleration is given by:
a = K/(L - x)^2
K = 100 m^3/s^2
L = 0.169 m
This acceleration will result in a force:
F = m * a
F = m * K/(L - x)^2
This force will perform a work:
W = F * L
The ball will advance only until x = L - D/2
[tex]W = m * K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]
This work will be converted to kinetic energy
W = Ek
Ek = 1/2 * m * v^2
[tex]1/2 * m * v^2 = m * K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]
[tex]1/2 * v^2 = K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]
First we solve thr integral:
[tex]K \int\limits^{L - D/2}_0 {\frac{1}{(L - x)^2}} \, dx[/tex]
We use the replacement
u = L - x
du = -dx
And the limits
When x = L - D/2, u = D2/2, and when x = 0, u = L
[tex]-K \int\limits^{D/2}_L {u^-2}} \, du[/tex]
K / u evaluated between L and D/2
2*K / D - K / L
Then
1/2 * v^2 = 2*K / D - K / L
1/2 * v^2 = K * (2/D - 1/L)
v^2 = 2*K*(2/D - 1/L)
[tex]v = \sqrt{2*K*(2/D - 1/L)}[/tex]
[tex]v = \sqrt{2*100*(2/0.0025 - 1/0.169)} = 398 m/s[/tex]
