Momentum question.

  • #1
Guys I'm having a little problem in undersatnding a few things:
We have a wall and a ball of mass 'm'. The ball is thrown towards the wall with speed u and rebounds with speed 'v'. The coefficient of restitution is, e is 1/3. The question is to find the value of 'v'.
Thats how I think it should be solved:
mu=mv (since the final velocity of the wall is "CONSIDERED" zero)
=> u=v

Now if I apply newton's law of restitution and simplify:
v=-eu (putting this into u=v gives us)
u=-eu
cancelling u gives e=-1 but e=1/3. Where did I go wrong? I just cant understand it.
Any help will be greatly appreciated.
 
Last edited:

Answers and Replies

  • #2
2,076
2
How can v=u given than e=1/3? The energy need not always be spent in speeding up an object. v = -1/3u.
 
  • #3
Oh my mistake. It is v=-eu. Original post edited.

neutrino said:
How can v=u given than e=1/3?

I asked this question because nowadays we are doing oblique collisions between a wall and a ball in mathematics.
This is an example question(it appeared in exams a few years ago).
Velocity of the ball=u
angle which the ball makes with the smooth wall when it collides=45deg.
angle after collision=theta
velocity after collision=20m/s
e=1/3.
Find u and angle theta.
I've got the answer by considering the component of velocity parallel and perpendicular to the wall. What I want to know is why Law of conservation momentum gives us false result considering it momemtum is conserved in ALL collision.
 
  • #4
141
0
Lets say you throw the ball against the wall and no energy is lost whatsoever. You would expect the ball to hit the wall and bounce back. If the ball loses no energy, then you also expect that its final speed is equal to its initial speed (in the opposite direction). Now examine it's momentum. Initially it had

p_0 = +mv

while finally it has

P_1 = -mv

This is a net momentum "gain" of |2mv| for the ball in the direction anti-parrallel to its initial velocity. Where has the momentum gone (or come from)?

The answer is that the wall suffers a net "gain" of -|2mv| of momentum. If the wall is fixed then the gain in momentum will be manifest as phonons (vibrations) in the wall. This is why someone else on the other side of the wall would hear you throwing the ball against it.

So momentum is conserved in the entire collision. It's just not very useful for calculating the longitudinal component of momentum for your ball.
 

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