A marble ball of mass 1kg falls from the top a building of height 10m and penetrate 0.5m deep in the Sand calculate the average resistance offered by the Sand and time taken by the ball to penetrate.
Answers
Answer:196 N , 1/14 s
Explanation:First we find the final velocity attained by the object just before reaching the ground:
u = 0 m/s
a = 9.8 m/s²
s = 10 m
v = ?
Using v² - u² = 2as,
we derive v² = u² + 2as
v² = 0 + 2(9.8 m/s²)(10 m)
= 196
v = √ 196
= 14 m/s
Secondly, we find the acceleration the a happens when the object collides with sand. Therefore, the v we found becomes the u and the v that we shall take will be 0 as the object comes to rest.
u = 14 m/s
v = 0 m/s
s = 0.5 m
a = ?
Again we derive using v² - u² = 2as,
We derive a = (v² - u²)/2s
a = (0 - 14²)/(2 x 0.5m)
= -196/1 = -196 m/s²
Now that we know the acceleration of object A (the ball) with it's mass, we can calculate the resistance offered by the sand (object B).
Fab = ma
= 1 kg x -196 m/s²
= -196 N
According to Newton's Laws of motion,
Fab = -Fba
Henceforth, Fba = +196 N
So we know of the resistance.
To find the time, we use a = (v - u)/t.
We know that
a = -196 m/s²
u = 14 m/s
v = 0 m/s
t = ?
Using the formula for acceleration, we derive
t = (v - u) / a
= (0 - 14 m/s) / -196 m/s²
= (-14 m/s) / (-196 m/s²)
= 1/14 s