Aball falls for 9.0 s increasing its kinetic energy by 2700. if the force acting on the ball in 6.0 n find the following quantities:

change in velocity:
average velocity:
initial velocity:
final velocity:
initial momentum:
final momentum:

Change in velocity:  88 m/s

Average velocity:  50 m/s

initial velocity:  5.9 m/s

Final velocity:  94 m/s

Initial momentum:  3.6 kg m/s

Final momentum: 58 kg m/s

Explanation:

Acceleration = change in velocity / time

9.8 m/s² = Δv / 9.0 s

Δv = 88 m/s

Work = change in energy

Fd = ΔE

(6.0 N) d = 2700 J

d = 450 m

Average velocity = distance / time

v_avg = 450 m / 9.0 s

v_avg = 50 m/s

v − v₀ = 88 m/s

½ (v + v₀) = 50 m/s

Solving the system of equations:

v + v₀ = 100 m/s

2v = 188 m/s

v = 94 m/s

v₀ = 5.9 m/s

Use Newton's second law to find the mass:

F = ma

6.0 N = m (9.8 m/s²)

m = 0.61 kg

Find the momentums:

p₀ = (0.61 kg) (5.9 m/s) = 3.6 kg m/s

p = (0.61 kg) (94 m/s) = 58 kg m/s

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s