# Collision of two Balls

1. Oct 6, 2007

### sam.

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

Ball 1 (B1) and Ball 2 (B2) are located at (x,y)=(0,0) and (x,y)=(d,h).
At t=0, B1 is sent towards the initial location of B2 with a speed vi. At the same
instant that B1 is launched, B2 falls towards the ground with zero initial velocity.
Assume there is no air resistance.

A diagram is attached below.

1. When and where do B1 and B2 collide?
2. If the initial speed of B1 is larger than vi, does a collision occur?
3. If B1 is directed towards a point slightly above the initial location of B2, can a
collision occur?
4. If B2 has an initial speed Vi in the negative y-direction, can B1 collide with B2?

2. Relevant equations

1. v_f = v_i + a$$\Delta$$t
2. s_f = s_i +v$$\Delta$$t +1/2a$$\Delta$$t^2
3. v_f^2 = v_i^2 +2a$$\Delta$$s
4. s_f = s_i + v$$\Delta$$t

3. The attempt at a solution

I solved for the distance that Ball 1 travels using equation 4 and got:
s(ball 1) = cos$$\theta$$v_it_1
Then for Ball 2 I used equation 2:
S(BALL 2) = h + 4.9t_1^2
Then I made both these equal each other:
h + 4.9t_1^2 - cos$$\theta$$v_it_1 = 0
Now I know I can solve for time using quadratic formula but I'm not sure how to find where the balls meet. Also I am completely lost on how to solve for the rest of the questions. Please, someone help me out!! Any help is appreciated!

File size:
151 KB
Views:
84
2. Oct 8, 2007

### physics2004

you can't use equation four because it is not constant velocity, (gravity has a role) therefore you'd have to use equation two twice for B1 and B2 then have the two equal because that will give you time of collision (it won't be a quadratic) then use T to get distance. Btw you used Cos(theta) when its Sin because its vertical component. and you know the x distance is d because thats where B2 is dropped so use vertical distance of B1 and d to find co-ordinate of the collision. i did it just now everything worked out fine.