Calculating Velocity & Retarding Force of Dropped Mass

AI Thread Summary
The discussion focuses on calculating the velocity and retarding force of a metal ball dropped from a height of 90 meters. The initial velocity upon impact was calculated to be 42 m/s using the SUVAT equation. For the retarding force, it was clarified that simply using F=ma with gravitational acceleration is incorrect, as the ball experiences additional forces while penetrating the sand. Instead, the required deceleration must be calculated using the SUVAT equations based on the stopping distance of 6.0 cm. The net force will be greater than the gravitational force due to the short stopping distance, indicating a significant retarding force from the sand.
Apothem
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Homework Statement



"A metal ball of mass 0.50 kg is dropped from the top of a vertical cliff of height 90m. When it hits the beach below it penetrates at a depth of 6.0cm. Calculate:
a) the velocity of the ball just as it hits the sand
b) the (average) retarding force of the sand

Homework Equations



SUVAT equation: v2=u2+2as
Newton's Second Law: F=ma

The Attempt at a Solution



For part a) I used suvat (s=90m, u=0ms-1, v=v ,a=9.81ms-2 , t=t) and used the suvat equation to calculate the final velocity as 42ms-1

For part b) I am unsure, do I just use F=ma (with m=0.50kg and a=9.81ms-2) to get an answer of 4.91N, or am I missing something?
 
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Apothem said:

Homework Statement



"A metal ball of mass 0.50 kg is dropped from the top of a vertical cliff of height 90m. When it hits the beach below it penetrates at a depth of 6.0cm. Calculate:
a) the velocity of the ball just as it hits the sand
b) the (average) retarding force of the sand

Homework Equations



SUVAT equation: v2=u2+2as
Newton's Second Law: F=ma

The Attempt at a Solution



For part a) I used suvat (s=90m, u=0ms-1, v=v ,a=9.81ms-2 , t=t) and used the suvat equation to calculate the final velocity as 42ms-1

For part b) I am unsure, do I just use F=ma (with m=0.50kg and a=9.81ms-2) to get an answer of 4.91N, or am I missing something?

For part b, no, it's not just the case that a = g. That would be true only if the object were in free fall (meaning gravity is the only force acting). Once it hits the ground and starts burrowing, this is no longer true. There is another force (from the ground).

What you have to do is apply the same equation as you did for part a, this time with u = the v you calculated from part a, and v = 0 since it comes to rest. From this you can figure out what acceleration is *required* in order to slow the object down over that distance. This required acceleration tells you the net force.
 
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Apothem said:

Homework Statement



"A metal ball of mass 0.50 kg is dropped from the top of a vertical cliff of height 90m. When it hits the beach below it penetrates at a depth of 6.0cm. Calculate:
a) the velocity of the ball just as it hits the sand
b) the (average) retarding force of the sand

Homework Equations



SUVAT equation: v2=u2+2as
Newton's Second Law: F=ma

The Attempt at a Solution



For part a) I used suvat (s=90m, u=0ms-1, v=v ,a=9.81ms-2 , t=t) and used the suvat equation to calculate the final velocity as 42ms-1

For part b) I am unsure, do I just use F=ma (with m=0.50kg and a=9.81ms-2) to get an answer of 4.91N, or am I missing something?

Part a) is alright .

For Part b) the ball starts with an initial velocity 42ms-1 and ends up being at rest covering a distance of 6.0 cm .During the motion the ball is under the influence of two forces,force of gravity and the retarding force of the sand .Using F=Ma with a=9.81 is incorrect. Value of 'a' can be calculated using SUVAT equations .Now using F=Ma will give you the net force .
 
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You should find that a > g due to the very short stopping distance.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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