When you practice tennis by serving balls against a brick wall your typical serv

Question

When you practice tennis by serving balls against a brick wall your typical serve hits the wall at 30 m/s. The coefficient of restitution of the tennis ball is 0.75. How fast is the ball going when it first bounces back from the collision with the wall?

Hanging a fish that weighs 9.8 N from your Acme De-Liar Deluxe spring scale causes the bottom of the spring to move down 0.02 m. You had no luck when you were fishing in the James River just behind the CNU campus, so you went to Harris-Teeter and bought a 2kg fish to bring home for dinner. How far will that fish cause the bottom of the spring scale to move down when you hang it from the scale? (Assume the same spring constant)

If the earth doubled in mass but stayed the same size, what would be the acceleration due to gravity at the surface? 1 m/s2 The dimensions of the earth as we currently know it are: Mass = 5.98E24 kg, Radius = 6.38E6 m and the universal gravitational constant = 6.67e-11 nm2/kg2

Explanation / Answer

1. speed of ball on return = 0.75 x 30 = 22.5m/s

KE = 1/2mv^2

KEbefore = 1/2 m 30^2
KE after = 1/2 m 22.5^2

Fraction of KE remaining = 22.5^2/30^2 = 0.56 or 56%

2. Ok you use the equation F=kx where F equals the force k equals a spring constant and x equals the distance the spring moved.

So first you plug in the numbers for your first fish to get your spring constant.

9.8 N =k *.02 m
9.8 N /.02 m=k
490 N/m^2 = k

so then you use your constant for your spring (k) and your force (2*9.8) for your second fish. (the 2 times 9.8 comes from the weight of the fish)

19.6 N = 490 N/m^2 * x
19.6 N / 490 N/m^2 =x
x=0.04 m

3.

The depends if you are asking about the acceleration of the mass towards another body, or if you are asking about the acceleration of another body towards the mass.

Assuming the former:

F(g) = G * m1 * m2 / r

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