Calculating Velocity After Impact: What Went Wrong?

  • Thread starter Thread starter DarkFalz
  • Start date Start date
  • Tags Tags
    Velocity Wall
AI Thread Summary
The discussion revolves around the calculation of velocity after a ball impacts a wall. The initial momentum of the ball is correctly calculated, but confusion arises when determining the final velocity of the ball after the collision. It is clarified that while the wall's mass is significant, its velocity is effectively zero, leading to the conclusion that the ball should rebound with a velocity of -10 m/s. The conversation emphasizes the need to apply both momentum and energy conservation equations to accurately solve for the velocities involved. Ultimately, the importance of clearly stating equations and understanding the principles of elastic collisions is highlighted.
DarkFalz
Messages
71
Reaction score
0
Hello,

i have the following questions. Let's assume that i throw a ball with mass 1kg and velocity 10 m/s at a wall with mass 100000000000000000 and velocity 0

using linear momentum i have in the beginning i have p = 1*10+ 100000000000000*0 = 10

after the impact the wall stays still and so the ball should go back with the same speed but the opposite direction, so i should have

Vf = - Vi

but if i do the calculations:

10 = 1*vf + 100000000000000*0 <=> 10 = vf <=> vf = 10

what have i done wrong? the speed should be -10
 
Physics news on Phys.org
The velocity of the wall after collision is not exactly zero. Even with a small velocity, the product of the mass and the velocity is not negligible.

You should solve it this way:

m: mass of the ball
M: mass of the wall
v0: the velocity of the ball before collision
v: the velocity of the ball after collision
V: the velocity of the wall after collision


mv_{0}=mv+MV


This equation is true for both elastic and inelastic collisions but it has two unknowns. We need another equation to find the velocities. That is the conservation of energy. For an elastic collision, the kinetic energy is conserved:

\frac{1}{2}mv_{0}^{2}=\frac{1}{2}mv^{2}+\frac{1}{2}MV^{2}

from the two equations, you obtain

v=\frac{m-M}{m+M}v_{0}

and

V=\frac{2m}{m+M}v_{0}
 
Last edited:
You need to clarify the question, in particular the line after "but if i do the calculations:" is not clear. Try writing the equation you're using in its general form (using letters) then in the line directly under it, repeat the equation with the specific numbers filled in, and use units.

It looks like what you're trying to say is,
initial momentum of the ball = negative the final momentum of the ball (the wall is irrelevant since it's always multiplied by zero)
pf=-pi
mfvf=-mivi
1 kg x vf=-1 kg x 10 m/s
So vf=-10 m/s.
Now, what's the question?
 
Thread 'Is 'Velocity of Transport' a Recognized Term in English Mechanics Literature?'
Here are two fragments from Banach's monograph in Mechanics I have never seen the term <<velocity of transport>> in English texts. Actually I have never seen this term being named somehow in English. This term has a name in Russian books. I looked through the original Banach's text in Polish and there is a Polish name for this term. It is a little bit surprising that the Polish name differs from the Russian one and also differs from this English translation. My question is: Is there...
I know that mass does not affect the acceleration in a simple pendulum undergoing SHM, but how does the mass on the spring that makes up the elastic pendulum affect its acceleration? Certainly, there must be a change due to the displacement from equilibrium caused by each differing mass? I am talking about finding the acceleration at a specific time on each trial with different masses and comparing them. How would they compare and why?
Back
Top