Height at which two balls collide traveling opposite direcitons

  • Thread starter Thread starter Tachypsychia
  • Start date Start date
  • Tags Tags
    Balls Height
AI Thread Summary
The discussion revolves around a physics problem involving two balls colliding while traveling in opposite directions. Ball A is dropped from a height h, while Ball B is thrown upward from the ground, with the condition that Ball A's speed is twice that of Ball B at the moment of collision. Participants emphasize the importance of using kinematic equations to find the height of the collision, suggesting that the initial velocity of Ball B can be treated as a parameter that will ultimately cancel out. After working through the equations, it is concluded that the collision occurs at a height of 2/3h of the building's total height. The conversation highlights the value of solving problems symbolically before substituting numerical values.
Tachypsychia
Messages
2
Reaction score
0

Homework Statement


Ball A is dropped from the top of a building of height h at the same time that Ball B is thrown vertically upward from the ground. When the balls collide, they are moving in opposite directions, and the speed of A is twice the speed of B. At what height does the collision occur?


Homework Equations


Kinematic equations:
Vf=vi+ at
xf=xi + 1/2(vi+vf)t
xf=xi+vi*t+1/2at^2
vf^2=vi^2+2a(xf-xi)

V(ball A) = 2V(ball B)


The Attempt at a Solution



I've done kinematics equations before, but it's been a while and there is a little more involved with this one, as you're looking for the moment of impact's height where the velocity of falling ball A is twice the speed of rising ball B.

I know the velocity of ball A is simply V(Ball A) = -9.8m/s^2*t
Ball B has a velocity of V(ball B) = V-9.8m/s^2*t

The collision must occur before or at the maximum height Ball B can reach with its initial velocity, but since I can't solve for it, I'm not sure what to do.

I'm fairly sure I can set the values equal to each other because of the relationship of V(ball A) to V(ball B) but I'm having trouble conceptualizing, and don't know if they even want an exact or symbolic answer.

V(ball A) = -9.8*t
V(ball B) = (-9.8*t)/2

I need to solve for time to get height, and height to get time. But I don't have any values for velocity. The only value I have is the constant of acceleration and I feel like that isn't enough to work this out.
 
Physics news on Phys.org
You did the velocities well. Now you need to write equation for the height where the balls meet. They have to be at the same spot at the same time.

Use the initial velocity of B as parameter. It will cancel out. One more piece of advice: use the symbol g, do not plug in the number too early.

ehild
 
I knew that there was something I was missing out on! After solving for final height for ball a and ball b and setting them equal to each other, it came out to be a symbolic answer: 2/3h of the building. Thanks ehild! I got some good tips for conceptualizing these problems the other day, and not pluging in known values until you've solved algebraically was one of them. =)
 
Good work!:smile:

ehild
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top