Explanation of acceleration of a ball bouncing up and down on the ground

[city25]

I am from HK. Hope you guys can understand my poor English! ^^
actually, this is not for my homework but my preparation for the public exam.

Homework Statement

The situation is that a ball bouncing up and down on the ground in vertical direction.
The question is to choose which graphs best describes the variation of its acceleration a.
It is a MC-question. Althought I know the ans., I want to have a full explanation but I am not sure if I am correct.

Homework Equations

No

The Attempt at a Solution

We all know that there is a sudden change in acceleration to the opposite sign at the moment that the ball hits ground. I want to explain it. Please comment on my explanation.


　　u↓　↑v
　　　Ｏ　　　　↓+ve
－－－－－－－
　　ground

where u ≥ -v
ps. the ball with mass m.Let's consider momentum.
impact force
= (mv - mu)/t
≥ [m(-u) - mu]/t (since v ≥ -u)
= -2mu/t which is negativenet Force = ma
impact force - mg = ma
ma ≤ -2mu/t - mg
a ≤ -2u/t - g

hence, when the ball hits the ground, it experiences acceleration in the sign opposite to gravitational field.

Am I correct?
Can I explain in other way(s)? or simplier way(s)?
I want explanation as much as possible.

 

[zhermes]

Well, for a perfect collision between an immovable ground and perfectly elastic ball, the acceleration would be infinite, and all you need are the conservation laws. For a non-ideal situation, you have to consider the ball as some sort of spring, you can get an expression for the time the impulse will be delivered based on the spring constant (or equation).

The rate at which the acceleration changes direction--which, I think, is what you want to find out--depends sensitively on how you're modeling/thinking about the ball.

 

[city25]

Well, for a perfect collision between an immovable ground and perfectly elastic ball, the acceleration would be infinite, and all you need are the conservation laws. For a non-ideal situation, you have to consider the ball as some sort of spring, you can get an expression for the time the impulse will be delivered based on the spring constant (or equation).

The rate at which the acceleration changes direction--which, I think, is what you want to find out--depends sensitively on how you're modeling/thinking about the ball.

So, you mean that there is different explanation when it is in different cases?
Is there any wrong or improper concept?

How about this?
Since t ∝ kx where k is the spring constant and x is displacement (how should I difine x?)
=> t = μkx where μ is a constant.

hence, a ≤ -2u/μkx - g
since x change from -ve to +ve, and then -ve, acceleration changes its direction "gradually".

is it correct?

 

[city25]

oh, i must make a mistake

now, let v be the velocity of the ball when it hits ground
impluse
= (mv - mu)/t

net Force = ma
impluse - mg = ma
ma = (mv - mu)/t - mg
a = (v - u)/t - g

since t ∝ 1/k => t = μ/k where k is spring constant, μ is a constant
hence, a = (v - u)k/μ - g
da/dt = (k/μ)(dv/dt) - uk/μ - g

When the ball hits ground,
speed of the ball ↓,
=> (dv/dt) is -ve.
=> da/dt is also -ve
=> slope of the a-t graph is -ve

when the ball leaves ground,
speed of the ball ↑,
=> (dv/dt) is +ve.
=> da/dt may become +ve
=> slope of the a-t graph may be +ve

however, it seems that i can't say "may be +ve" in this case, i should say "must be +ve". i don't know whether i made something wrong before. If no, how should I deduce "must"?

 

[rwisz]

Your explanation is correct. Another way to think about it is through the concept of impulse and momentum. When the ball hits the ground, it experiences a sudden change in momentum due to the impact force. This change in momentum results in a change in velocity and therefore a change in acceleration. Since the impact force is in the opposite direction of the ball's initial velocity, the acceleration is also in the opposite direction. This is why we see a sudden change in acceleration when the ball hits the ground.