A0.060 kg ball hits the ground with a speed of -32m/s. the ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 n force on the ball. what is the magnitude of the velocity after it hits the ground?

a. 9.3m/sb.12m/sc.41m/sd.73m/s

A:

Impulse is given by:

FΔt = Δp

We know the mass and velocity and we need the final velocity so lets break this down and isolate

FΔt = mv(final) - mv(initial)

Factor out m

FΔt = m(v(final) - v(initial))

FΔt / m = v(final) - v(initial)

Add v initial

(FΔt / m) + v(initial) = v(final)

(55x45ms / 0.060) - 32 = v(final)

v(final) = +9.25 m/s or 9.3

d eras and periods

what experimental evidence led scientists to change from the previous model to this one?

a few of the positive particles aimed at a gold foil seemed to bounce back.

the colors of light emitted from heated atoms had very specific energies.

experiments with water vapor showed that elements combine in specific proportions.

cathode rays were bent in the same way whenever a magnet was brought near them.