Physics homework problem - question

[rafalski]

Hello, I have one number left in my physics homework which I can't figure out

Question:

A ball of 300g hits the floor at a velocity of 2,50 m/s at an angle of 60* relative to the vertical. The vertical force in function with time between the floor and the ball is:

from 0 to 50 N : from 0 to 1 ms.
from 50 N to 100 N : from 1 to 2 ms.
constant 100 N : from 2 to 3 ms.
100 to 50 N : from 3 to 4 ms.
and 50 N to 0 N : from 4 to 5 ms.

What is the velocity of the ball after the collision if the ball bounces? Give the answer in terms of unit vectors and polar notation.

So I know that I have to find the area under the graph which will give me the impulsion. From the formula I = m(V - Vo) I can get my velocity but
what I can't figure out is how to write it in unit vector and polar notation, because supposedly the impulse is 0 in the horizontal direction.

Thank you for your time

[rafalski]

would my velocity right before the ball hits the ground be:

(2,50 cos70 i ) m/s + ( 2.50 sin70 j ) m/s?

Then with the impulsion I get from the area of the graph I can find the velocity,
which would be:

I = (0.005 s. + 0.001 s.)(100N) / 2
I = 0.3 N*s

m(V - Vo) = 0.3 N*s
V = (0.3 N*s + mVo) / m = 0.3 N*s + (0.250kg)(2.50 m/s) / 0.250kg = 0.75 m/s

Now how would I express that in unit vector? From unit vectors I know that I can get the polar notation by using tan0.

[Ibid]

You're most of the way there, a couple minor fixes, which you actually almost caught yourself: The force affects ONLY the vertical velocity. Impulse is, as you noted, split into two components. The vertical (incidentally, that should be 0.002s + 0.001s in that calculation, those triangles on the graph are each 2ms, and the centre is 1ms), and the horizontal impulse is 0. I'm going to use subscripts for the directions, i.e., v_j is vertical velocity.

So, what you calculated is I_j
As you noted, the horizontal impulse I_i is zero.

The vertical velocity change equation is I_j = m(V_j - V_j0)

The horizontal velocity change is I_i = 0 = m(V_i - V_i0)

V_j = 2.50m/s * cos60

V_i = 2.50m/s * sin60

The importance of that is that when you're using the vertical impulse, it will only affect the vertical velocity V_j, V_i is only going to be affected by horizontal impulse (here zero).

The second thing to watch for is to remember which way your negative is. If V_j starts as being positive, then the j direction is DOWN, not up, and a negative V_j implies a ball that is moving upwards.

EDIT: Oh, I may have swapped the directions on you, I was using j as the vertical, but you can swap i and j around as long as you're consistent.

[rafalski]

thanks a lot! I see I made a mistake and wrote an angle of 70 and 0.250kg, those were values from another problem, sorry!