# Mass and stopping distance

1. Aug 27, 2015

### Viraam

1. The problem statement, all variables and given/known data

2. Relevant equations
$F = \frac{m(v-u)}{t}$

3. The attempt at a solution
The mass of cricket ball is greater than mass of plastic ball.
Change of velocity is same for both. Now since the rate of change of momentum of cricket ball is greater the force needed to bring it to stop is greater. Since $F \alpha \frac{1}{t}$ the time taken by the cricket ball to stop is small. Therefore the plastic ball will cover greater distance.

Is my answer correct. Some sites claim that the cricket ball will cover larger distance. Is that true, Please answer ASAP

2. Aug 27, 2015

### andrewkirk

That proportionality is only correct where the other variables remain constant. But $m$ differs between the two cases, so the proportionality does not hold.
The problem is not solvable without more information about what you are supposed to take into account, in particular air resistance and rolling resistance, the size of the two balls, whether the plastic ball is solid, and so on.
In general one would expect the cricket ball to roll further than a hollow plastic ball because, if they are the same size, they'll encounter the same air resistance but the cricket ball will have more initial momentum and KE. But without that extra info, the problem is not solvable.

3. Aug 27, 2015

### Viraam

Why is it that the cricket ball will go farther distance than the plastic ball. Please assume that the plastic ball is hollow and lighter than the cricket ball. Please explain in terms of $F=\frac{m(v-u)}{t}$ The $v-u$ is equal for both the balls only mass $m$ is differing. Therefore we can say that a larger force will be needed to stop the cricket ball because of which it travels a larger distance. Is this correct?

4. Aug 27, 2015

### NickAtNight

I am leaning towards the plastic ball going farther.

Why are we making this assumption? Is this an assumption provided by the provider of the problem, or one you are making?

A cricket ball is well defined by rules. The plastic ball is open to your imagination the way this problem is stated..

Why did you pick this equation? What Force(s) work on the balls to slow them down and cause them to stop?

I do not think so.

5. Aug 27, 2015

### NickAtNight

6. Aug 27, 2015

### andrewkirk

That equation is not correct for this case because it assumes that force is constant over time. In this case it will not be constant because air resistance increases with velocity. A correct version of the equation is $v-u=\int_{t_1}^{t_2} F(t)dt$.

We can approximately ignore that problem if we replace F in your equation by the average force over time. Then it will be true that, to stop the cricket ball in the same time as a much lighter plastic ball will require a greater average force. Conversely, because the forces will not be very different between the two (Air resistance is roughly similar. The cricket ball will suffer greater rolling resistance than the plastic ball because of its greater weight but if it's rolling on a smooth surface then I think that will be much less than the air resistance), the cricket ball will take a longer time to stop, and hence will travel further.

7. Aug 28, 2015

### NickAtNight

How are you doing. Did you figure out the correct equation yet?

8. Aug 28, 2015

### Viraam

Actually, this question is appearing in our Chapter called Force and motion where the following equations are given:
$F=ma \\ F= \frac{m(v-u)}{t} \\ v = u+at \\ v^2 - u^2 = 2as \\ s = ut + \frac{1}{2} at^2 \\ \text{Conservation of Linear Momentum}$

9. Aug 28, 2015

### Viraam

Actually, this question is appearing in our Chapter called Force and motion where the following equations are given:
$F=ma \\ F= \frac{m(v-u)}{t} \\ v = u+at \\ v^2 - u^2 = 2as \\ s = ut + \frac{1}{2} at^2 \\ \text{Conservation of Linear Momentum}$
It would be great if you could give me an answer with regard to this

10. Aug 28, 2015

### andrewkirk

All of those equations except the first are for a force that is constant over time, which will not be the case for this question. So they are not applicable.
Given the lack of information provided in the question, the best answer you can give will be based on the broad facts outlined in my second para in post 2. The cricket ball will win if the plastic ball is the same size as the cricket ball and the floor is polished wood. Why? Because air resistance is the dominant force and it's approx the same for both balls. So, given the mass of the cricket ball is greater, its stopping time will be longer and so its distance will be further.

11. Aug 28, 2015

### NickAtNight

We are not allowed to just give the answer. :(

1) You know that both balls have an initial velocity of $v_i$
2) Look more closely at the left side of the equation. The F is actually the sum of all the Forces acting on the object. $\sum F$
3) The initial force that started the motion (your hand) is gone.
4) there are two Forces (mentioned by the other poster) working pro slow the ball.
5) What are the formulas for those two forces?
6) What properties affect those two forces.
Hint: the other poster mentioned size of the ball. Skydivers change the rate that they fall at by changing their shape to increase/decrease their size and increase/decrease the wind resistance.
Hint: the web page I posted previously gives a formula for the other force.
7) it is a logic problem, so the answer is the possible answers depending on the circumstances. Not saying that well. But something like A goes further if x and y are more important. And B goes further if z and a are dominant.

12. Aug 28, 2015

### CWatters

I think some of you are over thinking the problem.

Cricket balls are usually heavier than plastic balls of the same size.
Both start with the same velocity so one has more initial KE and momentum.
In the absence of info it would be reasonable to assume both encounter the same resistance (whatever the source might be).

13. Aug 28, 2015

### Viraam

4) Air Resistance and Rolling Friction
5) Not Sure. The formula for rolling friction in the site is $F_R = μ_RWcos(a)$
6) Friction: Irregularities between the surfaces. Air Resistance: Shape of the body.

14. Aug 28, 2015

### NickAtNight

Let's assume air resistance is negligible.

The only force to stop the balls is now rolling friction.
... Since we are on a flat surface, we can drop the cos(a). So $F_r = μ * W$

So since the balls have the same initial velocity and F=m*a. Then F/m=a. So does the mass term on both sides of the equation cancel leaving us with just the rolling friction coefficient.

So if you have a plastic ball, of same mass and surface area as the cricket ball, then the friction coefficient is the determining factor for which ball stops later.

Now there are all kinds of plastics out there. So depending on how this ball is made and its resulting rolling friction coefficient, sets the answer.

A nice smooth, hard plastic ball would probably have a lower friction coefficient and roll a longer distance than a cricket ball.
... One of those super elastic rubber plastic bouncing balls, softer, but with energy recovery from deformation, and smooth, would also probably go further.
.... But a squishy plastic, with a deliberately made rough surface - as long as it is rougher than the cricket balls surface, would go less distance.

15. Aug 28, 2015

### NickAtNight

Now add back in air resistance...

Cross sectional area is important. As is the skin friction. What can you say about this force?

https://en.m.wikipedia.org/wiki/Air_resistance

16. Aug 28, 2015

### andrewkirk

I do not think it would be reasonable. As I understand it, rolling resistance is proportional to weight, so the retarding force of rolling resistance would be greater for the cricket ball. So, ignoring rotational effects, it seems plausible that, in a vacuum, both balls may take roughly the same distance to stop. In fact it even seems plausible that the plastic ball would take longer to stop if hollow (as plastic balls typically are), because its ratio of moment of inertia to mass will be higher than for the solid cricket ball.

It appears to be only the air resistance that provides the strong differential effect that makes the cricket ball travel further.

17. Aug 28, 2015

### NickAtNight

Interesting. Learning a lot about cricket balls.

What will the prominent stitching do to the cricket balls friction?
Are we talking anew cricket ball or a used one. (Ignore me !)
I was looking for the density of the cricket ball. No luck so far. Quotes are from http://www.decathlon.co.uk/blog/cricket/everything-you-need-to-know-about-cricket-balls/ [Broken][/quote][/quote]

Last edited by a moderator: May 7, 2017
18. Aug 28, 2015

### NickAtNight

Let's go with a professional plastic ball... Hmmm, what would that be. How about a bowling ball !!! https://en.m.wikipedia.org/wiki/Bowling_ball

Is a cricket ball more or less dense than a plastic bowling ball?
In a test of a cricket ball versus a 16lb plastic bowling ball, the bowling ball would probably win hands down. :)

Our logic problem here did not limit us on mass or volume of the plastic ball - so we cheat and use a massive bowling ball.

Control for volume and make a mini urethane ball, and it probably still wins.

All kinds of cool information on the tricks professional bowlers got up to to enhance their score... https://en.m.wikipedia.org/wiki/Bowling_ball

Note: I have an old friend who's kid was the third youngest to score a perfect 300 in bowling. That is about 3x better than my typical score. They told me that bowling alleys oil their lanes differently for different games. One popular pattern is a Christmas tree shape. The location of the oil sets where location where the balls skid and where they roll. Once the ball stops skidding and starts to roll, then the spin the bowler put on the ball comes into play causing the ball to hook.

19. Aug 28, 2015

### NickAtNight

So did they explain where these equations come from? I suppose they just presented them.

So you have been using: $$\sum_{net}F = F_1 + F_2 + ... + F_i = m \times a$$

Now acceleration (a) is the rate of of change in velocity with respect to time.: $$a = \frac {\delta V} {\delta Time} = \frac {V_2 - V_1} {T_2 - T_1}$$

An object with zero initial velocity is accelerated at 10 m/s^2 for 10 seconds. What is the final velocity?
.....$$a = \frac {\delta V} {\delta Time} = 10 m/s^2= \frac {V_2 - 0} {10 - 0}$$
Rearrange to solve for V_2
......$$V_2 = 10 m / s^2 * 10 s= 100 m/s$$

graphical solution:
On graph paper, Draw a graph of gravitational acceleration (use 10 m/s^2 for simpler math) versus time for time from zero to ten seconds. It is a nice flat line. The change in velocity is simply the shaded area under the curve ! So count off the number of boxes for each second.

In the first second, we have 10 boxes, so we are at 10 m/s after the first second.
The second second adds ten more boxes, so we are now at 20 m/s.
The third second adds ten more, so 30 m/s
4,5,6,7,8,9. Are 40, 50, 60, 70 , 80, and 90 m/s respectively.
At 10 seconds, we will have 100 squares, and thus be at 100 m/s.

On graph paper, draw the velocity as a function of time for ten seconds. Notice that it is a nice straight line with a slope of 10 m/s.

Integration:
When we integrate, we are just adding up the area under the curve!

Position:
Now velocity (v) is just the rate of change in position with respect to time. $$v = \frac {\delta S} {\delta Time} = \frac {S_2 - S_1} {T_2 - T_1}$$

If the initial position was zero, what is the final position? We would need to use the average velocity (50 m/s is 1/2 for 100 and 0)
$$S_2 = 50 m/s * 10 seconds + o m = 500 m$$

Graphical solution:
Count up the squares underneath the curve (straight line)

So for 0 seconds, 0 m
1 second..... 0+ 10 = 10 m
2 seconds. .... 10 + 20 = 30 m
3 s.................... 30 + 30 = 60 m
...
10 s .............,,,,, 400 + 100 = 500 m

Plot your result position versus time on your graph paper. You should have a nice curve.

Note: if we go the other way, position to velocity to acceleration, we reverse the process. We 'differentiate'. No you are finding the slope of the line at each instant in time.
The slope of a squared line (position) is a straight line.
The slope of a straight line is a constant.

20. Aug 28, 2015

### NickAtNight

V (t) = \int_{0s}^{10s} a\, \delta t= [a \times t] + V_i = [10 \frac{m}{s^2} \times 10 s] + 0 = 100 \frac{m}{s}

S (t) = \int_{0s}^{10s} (at + c)\, \delta t = [\frac{1}{2} a \times t^2] + [V_i \times t] + S_i = [\frac{1}{2} * 10 \frac{m}{s^2} \times 100s^2] + [0 * 10 s] + 0 = 500 m