Ask Me Help Desk - Ping-pong or bowling.and momentum and jazz

a ping-pong ball and a bowling ball have equal momentum. They are stopped with the same force, in the same amount of time. Which goes further while being stopped? (distance) I want to say they are equal but am I right?:confused: the equations Vf^2=Vi^2+2ad and m=mv=Ft might help? Hope so, I don't want to fail this class and I am trying my best! Lol thank you.

Without thinking about it too much here's my thought.
The ping pong ball has much less mass than the bowling ball, therefore, if they have the same momentum the pingpong ball must have a much higher velocity. If they decelerate for the same amount of time the pingpong ball will travel farther.
Make sense?

Asterisk I believe you are halfway there.

They are not decelerated with the same deceleration, but with the same force, from F=ma we see that the deceleration is much greater for the lighter ping pong ball.

I do believe that the correct answer is that they both stop in the same distance. But it's too late for me to do the math for you :) I'll have a go in the morning.

A quick look will show that the ping-pong ball travels farther.

The magnitude of the acceleration is greater for the ping-pong ball, as its mass is smaller, from f=ma

Since the acceleration is greater, that means that the change in velocity in the stopping time will be greater. This means that the initial velocity of the ping-pong ball is higher. Since the deceleration is constant, the velocity is linear, and the average velocity for each is half of the initial velocity.

The distance is then calculated simply by multiplying the average velocity by the time needed to stop. Since the time is the same for both, the one with the higher average velocity goes farther.

Edit: Solving it out to verify. 1 will indicate the ping-pong ball, and two will indicate the bowling ball.

$f=ma$
$f{\small_1}=f{\small_2}$

Therefore,

$m{\small_1}a{\small_1} = m{\small_2}a{\small_2}$

Since

$m{\small_1}<m{\small_2}$

$a{\small_1}>m{\small_2} (1)$

So the acceleration of the ping-pong ball is larger in magnitude.
Bearing in mind that acceleration is negative here,

$v{\small_i}=v{\small_f}-at$

Since the final velocity is 0,

$v{\small_i}=-at$

We can find from (1) above that

$v{\small_{i1}}>v{\small_{i2}} (2)$

Acceleration here is linear, and final velocity is 0, so the average velocity is given by

$v{\small_{avg}}=\frac{v{\small_i}}2 (3)$

Distance traveled is simply

$d=v{\small_{avg}}t$

t is constant, so

$\frac{d{\small_1}}{v{\small_{avg1}}}=\frac{d{\smal l_2}}{v{\small_{avg2}}}$

Combining this with (2) and (3) above, we get that d is proportional to v, which leads to

$d{\small_1}>d{\small_2}$

yeah. :)

also, imagine the velocity vs time graphs for each. The x intercepts will be equal, they'll both be straight lines with negative slope and y intercept will be initial velocity. The area under the curve for the velocity vs time graph is equal to the distance traveled. So the area of each will be a triangle. Each triangle's base will be the same (same time to decelerate) and the ping pong ball has a taller height because it has a larger initial velocity, therefore it's area will be greater, therefore its distance will be greater.

Thank you all so much!! Would you help me with my fitch barrier question now, lol :p