Calculating Velocity & Retarding Force of Dropped Mass

[Apothem]

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?

 

[cepheid]

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.

 

[Tanya Sharma]

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 .

 

[CWatters]

You should find that a > g due to the very short stopping distance.

 

[Nickie]

I would say that your approach for part a) is correct. You used the appropriate equation and input the given values to calculate the final velocity of the ball. However, for part b), you are indeed missing something. The retarding force of the sand is not simply the force of gravity (9.81N) acting on the ball. In this scenario, the sand exerts a force on the ball that opposes its motion and causes it to decelerate as it penetrates the sand. This force is known as the retarding force. To calculate it, you can use the formula F=ma, where m is the mass of the ball and a is the deceleration due to the retarding force. The deceleration can be calculated by using the SUVAT equation with the final velocity of the ball (from part a) and the distance the ball penetrates into the sand (6.0cm or 0.06m). Once you have the deceleration, you can plug it into F=ma to calculate the retarding force. This force will be less than the force of gravity acting on the ball, as the sand is not strong enough to completely stop the ball's motion. It is important to note that this calculation assumes that the sand exerts a constant retarding force on the ball throughout its penetration. In reality, the force may vary depending on the depth of penetration and other factors.