(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

2. Relevant equations

[tex]\vec{p}[/tex] = [tex]m\vec{v}[/tex]

[tex]F_{net}[/tex] [tex]\times[/tex] ([tex]t_{2}[/tex] - [tex]t_{1}[/tex] )= [tex]m_{2}[/tex] [tex]v_{2}[/tex] - [tex]m_{1}[/tex] [tex]v_{1}[/tex]

3. The attempt at a solution

So part A was easy, All i did was find the Velocity just before it hits using constant acceleration formula and times that velocity by mass.

So, my attempt went something like this, since [tex]\vec{p}[/tex] = [tex]m\vec{v}[/tex] , I have mass I just need to find velocity just after it hits the table.

I tried using law of conservation of energy to find velocity, but I can only take the points when it hits the wall and not JUST AFTER because I don't know the height. So this didn't work out too well

I know I can find the force the ball exerts on the wall (and vice ver sa due to Newton's 3rd law), so because the ball was just released / drop, it's acceleration is just 9.8, that times the mass will give me the force when the ball hits the wall.Now CAN I assumethat the force of the ball when it's going back up is the same magnitude of the force just as the ball hits the ground on it's way down (because the wall and the ball remained in contact for .015 seconds)?

Now, if I was allowed to assume that, the force would be 4.9N (9.8 * .05).

Momentum Impulse theory states that net force times the difference in time = the difference in momentum,

[tex]F_{net}[/tex] [tex]\times[/tex] ([tex]t_{2}[/tex] - [tex]t_{1}[/tex] )= [tex]m_{2}[/tex] [tex]v_{2}[/tex] - [tex]m_{1}[/tex] [tex]v_{1}[/tex]

So I have Fnet, delta t, the masses, and [tex]v_{1}[/tex].

My second question would be, how do I select the 2 points so I can assign which one is t1 and which one is t2

again IF my understanding is right (please correct me), I am going to say t1 is when the ball JUST hits the wall and t2 its the point when the ball is about to leave contact with the wall, so t1 will be 0 and t2 will be .015 seconds.

So plugging all that back into my impulse momentum theory

I have

4.9 N (0.015s-0s) = [tex].05kg v_{2}[/tex] - .05kg * 0m/s (because it has JUST hit the wall and remained contact so no velocity)

Solve for [tex]v_{2}[/tex] I get 1.47m/s

multiply that number by its mass (.05kg)

and I get .0735 momentum, which they told me is wrong

So where did I go wrong / not understanding?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Momentum, Impulse and consevation of momentum?

**Physics Forums | Science Articles, Homework Help, Discussion**