Correct question:a 200 g ball is dropped from a height of 2m, bounces on a hard floor and rebounds to a height of 1.5m. What maximum force does the floor exert on the ball?
The diagram of the question is in the attachment.
Answer:
Explanation:
V=√2gh
g-10m/s²
let u be initial velocity ad v be final velocity,
u=√2*10*2
u=6.324m/s
v=√2*10*1.5
v=5.477m/s
from the diagram
t=5ms=0.005s
[tex]F=\frac{m(v-u)}{t}[/tex]
[tex]F=\frac{0.2(5.477-6.324)}{0.005}[/tex]
F=-33.87N (the negative shows direction)
From the diagram Fmax=2F
Fmax= 2*33.87
=67.74N
The maximum force exerted on the ball is 33600 N .
Initially, the ball has zero (u=0) speed, the ball is dropped from a height of
h = 2m.
From the third equation of motion under the influence of gravity:
v² = u² + 2gh
where v is the final speed
v = √2gh
v = √2×9.8×2 m/s
v = 6.26 m/s
Now after the rebound, the ball reaches a height h' = 1.5 m, at the final position the velocity is zero (v'=0), so its initial velocity after the rebound can be derived from below:
v'² = u'² - 2gh'
0 = u'² - 2×9.8×1.5
u' = 5.42 m/s
Time of impact , t = 5 ms = 0.005s
So, the average force exerted on the ball is the rate of change of momentum:
F = m(v - u') / t
F = 200(6.26 - 5.42) / 0.005 N
F = 33600 N
Learn more about laws of motion:
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