# Homework Help: When a ball is dropped

1. May 6, 2004

### indebluez

hi everybdy...
i noe how the vel against time graph looks like when a ball is dropped...
but i have no idea how the acceleration against time graph looks like....

i noe at the moment the ball hits the ground...the acceleration value is still constant...but what about when its fallin and riing again...how will it lloook?

the ans give by the book looks somethinh like this...

/-\/-\ as in the acc values increase linearly, then remains const..then falls linearly...and then same thing happens again...why is this???

plz help...plzzzzz

2. May 6, 2004

### Parth Dave

while the ball is going down it will have a constant acceleration of 9.8 m/s^2 due to gravity. However when it hits the ground the ball will compress and than it will push itself back into its normal position which will give it kinetic energy and cause it to go back upwards. It will start to lose kinetic energy as it is going back up until it eventually cannot go up anymore. At this point gravity again will start pulling it back down with a constant acceleration.

3. May 6, 2004

### Evgeny

The answer given by your book is a bit weird, I think: if gravity is the only force acting on the ball, then the acceleration should be constant throughout: -9.81 m/sec^2. But assuming something odd does happen which makes the acceleration change, your ball will first hit the ground at some constant acceleration, then due to it not being perfectly solid, it will compress (like parth said), then some of it's energy will be transformed into other things like heat, but some will still remain (plus the ground will push against the ball due to Newton's third law). Then it will continue bouncing, less and less energy being retained each time.
The acceleration therefore, assuming the initial bounce is in fact as the book describes it, is just going to be a set of values increasing linearly, but less and less increasing with each bonce.

hope this helps

4. May 7, 2004

### indebluez

oh i think i got it guys:)

when the ball is falling, acc is cont...and when the ball bounces off the floor...the acc is still const...but since takin downward motion as positive...the acc is positive...

but at the instant at which the ball hits the ground....the acceleration falls...
but parth dave why does the acceleration fall at the instant at which the ball hits the ground?

so i realise everytime at the instant the ball bounces...the acceleration falls...by less and less...

but why does the g fall to a neg value when the ball hits the ground? i cant see what the compression of the ball has anything to do with the acc...help plzzzz

and thanx for all the replies:)

5. May 7, 2004

### BobG

Acceleration due to gravity remains constant throughout, whether the ball is falling downward, bouncing off the ground, or moving upward.

However, gravity isn't the only source of acceleration. When the ball hits the ground, it applies a force to the ground due to the kinetic energy it has built up. Any particle (or object) applies a force to a second, an equal and opposite force is applied by the second onto the first. In other words, for an instant, you have an immediate acceleration in the opposite direction (upward) because of the interaction between the ball and the ground. The sum of that immediate upward acceleration and the constant downward acceleration due to gravity provides your overall acceleration for that instant.

In practice, both the ball and the ground will compress, meaning the opposite acceleration isn't instantaneous, but spread out over time. The compression and re-expansion also takes some energy, meaning the upward force and resulting acceleration won't quite be equal and opposite to the force the ball applied to the ground. In other words, the ball will bounce a little less high with each bounce.

6. May 7, 2004

### indebluez

hi:)
thanx for the reply:) it has cleared more stuff in my head...

so when the ball hits the table, theres an equal and opposite reaction from the table on the ball...asl well...

so then why is there an upward acceleration then? i jus cant see why theres an upward acc....if the forces exerted on each other is the same...

i understd the why the resultant acc is upward....but dont understnd why when theres an interaction betwn the ball n the table results in an upward acc as well..with g acting on the ball downwards...

thanx again!!:)

7. May 7, 2004

### BobG

First of all, saying the ball is accelerated towards the Earth is a simplification of what really happens. What is really happening is that two moving objects are colliding.

All objects have gravity. Two objects interacting are both pulled towards their combined center of gravity. The amount of acceleration each experiences from that force depends upon how much inertia (or mass) each object has. Since one of the objects is just a ball and the other is the Earth, I don't think the Earth is going to move enough for you to measure, which is why the simplified version works.

When the two objects collide, each object causes a change in the direction of the other object. If you were talking about two pool balls colliding, I think it would be easier to see why each causes acceleration (a change in velocity) in the other. In this case, the Earth isn't going to change much in a collision with a ball. The only object with a measurable change in velocity from this collision is going to be the ball.

Gravity's kind of a cool subject. Every object with any mass at all has a gravitational attraction on everything around it, no matter what the distance is. The force of gravity goes on forever - granted, a person doesn't have much mass and the force of gravity decreases with the square of the distance between two objects, so the gravitational force of one person is pretty darn small, but its there none the less. Every star you see in the sky is tugging on you just a little bit through the force of gravity. Likewise, every star you see in the sky will be affected at least a little by your own personal gravitational force.

Kind of reminds you that:

1) You're inescapably part of the universe and personally affected by it.

and

2) That's okay because you belong here and you affect every other single thing that exists in the universe with you.

8. May 7, 2004

### indebluez

hi bob! thank u so much...i get it now! for a ball bouncing on a table,

when the ball hits the table...what happens is that teh ball exerts an equal and opposite force on the table, that the table exerts on it....

but the acceleration tts experinced by the table and the ball is different....the acceleration experinced by the ball is far greater then the acc experinced by the table...and the accelertaion experinced by the ball is upwards coz no way its downwards as its not goin to go into the table....

but one thing i want to clear in my head is...whats the force thats exerted by the table on the ball...is that m(table)x acc(table) ????
and which direction is this....since its the force exerted by the table on the ball...should be upwards?

thanx again:)

Last edited: May 7, 2004
9. May 7, 2004

### BobG

The force causing the upward acceleration comes from the kinetic energy the ball gained as it sped up on its way down. So the amount of upward force would be related to the mass and velocity of the ball, not the table.

10. May 7, 2004

### indebluez

hi bob,
one last tihng....so the force exerted by the ball on the table is upwards?
and the force exerted by the table on the ball is downwards? i thought it was the oter way round....sorry i noe i sound really confused...but help!

coz the resultant acc is from the forces exerted by each body on each other right....or...isit the force of the bodies itself...instead of force exerted by each on each other....tts wkin here...

i think i got it...is it like since the ball is fallin...the rate of change of velocity is negative..so the accerleration of the ball is negative...so the force of the ball is upwards......and the force of the table is downwards....
and resrultant acc....is upwards on the ball

but as the ball bounces, the kinetic engery is going to fall...the vel is goin to fall...the acc of the ball will fall....but how come the resultant accerleration falls? shouldnt it increase....or when the ball bounces, is it becoz of the resultant acc of the (table and ball) or isit the acc of the ball itself....

again a million thanx

Last edited: May 8, 2004
11. May 8, 2004

### Staff: Mentor

During the time they interact, the ball and table exert equal and opposite forces on each other. (This from Newton's 3rd Law.) The table exerts an upward force on the ball; the ball exerts a downward force on the table.

The exact force that they exert on each other varies during the interaction and depends on lots of things, such as the kind of materials involved. (For example, a steel ball dropped on a steel table will create different forces than a soft rubber ball dropped on a rubber table.) But so what, no one will ever ask you for the details of the interaction. As was already explained, except for the brief time that the ball is in contact with the table, the only force on the ball is gravity pulling down.
I'm not sure what you are saying here, but you can always use Newton's 2nd law to calculate the net force on an object if you know its acceleration.
Again, I am not sure what you are saying here. But, as the ball is falling, the only force (ignoring air resistance) is gravity accelerating it downwards. During the interaction with the table, the table will exert an upward force: the net force on the ball will be upwards, causing an upward acceleration. (The force that the table exerts on the ball will vary from zero to a maximum upward force then back to zero during the collision.)
It's also true that the ball exerts an equal force downward on the table. But who cares what happens to the table?

12. May 8, 2004

### indebluez

but bobg said that the upward force is the acc and mass of the ball...but doc al says the force on the ball is the force exerted by the table on the ball.....so shouldnt it be mass n acc of the table?

13. May 9, 2004

### Staff: Mentor

Newton's 2nd law

I'm sure he just meant that the net force on the ball (which is upward during the collision) can be calculated via Newton's 2nd law as F = ma of the ball. This is always true.
During the collision there are two forces acting on the ball: the weight of the ball acting down, and the table pushing the ball upwards.
I think you are confused about the application of Newton's 2nd law. It always relates the net force on an object (the cause, if you like) to the resulting acceleration of the object (the "effect"). The mass X acceleration of the table will be equal to the net force on the table, not the ball.

14. May 11, 2004

### indebluez

hi doc al:)
okye i finally get it:) theres a net force actiing on the ball....this net force on the ball is the force that provides the acceleration upwards...
the net force on the ball, is the force of the table on the ball...right?
theres an equal n opp force on the table exerted by the ball...hence theres a force on the table as well.....
hmmm i hope i am rigt?
thanxxxx alot!

15. May 11, 2004

### Staff: Mentor

I think you've got it (close enough).
Right. During the collision, the net force on the ball is upwards.
"net" force means the (vector) sum of all the forces. On the ball we have two forces: its weight (acting down) plus the large force of the table on the ball (pushing up). The sum of those two forces gives a net force upwards.

It's the net force that determines what happens to the body--how it will accelerate. You can have high forces applied to a body and still have a zero net force. For example: step on a rubber ball, squashing it against the ground. There are several forces on the ball: your foot exerts a high force, the weight of the ball exerts a force, and the ground pushes up with a large force. But the net force is zero: they all cancel out.
Right. When the table exerts a force on the ball, the ball exerts an equal force on the table. That's Newton's 3rd law. (Note: this applies to the individual forces between bodies, not the net force.)