After collision the ball comes to rest, mass of the ball is?

AI Thread Summary
The discussion revolves around a physics problem involving an elastic collision between a ball and a rod on a frictionless surface. The key equations to solve the problem are the principles of conservation of mechanical energy and linear momentum. Initial attempts suggest that the mass of the ball should equal the mass of the rod (m = M), but the correct answer is m = M/4, which requires considering angular momentum due to the rod's rotation after the collision. Participants emphasize the need to account for both translational and rotational motion in their calculations. The conversation highlights the complexity of determining the instantaneous center of rotation during the collision.
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Homework Statement

:
A stick of length L and mass M lies on a frictionless horizontal surface on which it is free to move in any way. A ball of mass m moving with speed v collides elastically with the rod at one of its extreme end( as shown in the figure). If after collision ball comes to rest, what should be the mass of the ball?
a) m= 2M
b)m=M
c)m=M/2
d)m=M/4[/B]

I have attached a pic of the figure

Homework Equations

: Principle of conservation of mechanical energy(PCME) and principle of conservation of linear momentum(PCLM)
[/B]

The Attempt at a Solution

:
From PCLM,[/B]
mv= MV
(where V= final velocity of the rod)
From PCME,(as collision is elastic, we can conserve ME)
1/2mv2=1/2MV2
from these two equations I got m=M
But the book says m=M/4
I wonder if we need to conserve angular momentum..but can't figure out how will conservimg angular momentum change the answer
Thanks for your help
Sorry for any grammatical error
Sorry if its a duplicate thread
 

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The force given by the smaller ball during collision will also cause the rod to rotate. Thus, the rod will undergo a combined motion of translation and rotation. So conservation of angular momentum is to be used.
 
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Also, if the rod is rotating then its kinetic energy is not just ##\frac12 MV^2## where ##V## is the speed of its centre of mass.
 
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okay, So when I'm conserving the energy, that will include the rotational kinetic energy...right?
 
randomgamernerd said:
okay, So when I'm conserving the energy, that will include the rotational kinetic energy...right?
Yes.
 
Hello,
I have a doubt related to this problem. When the ball strikes the rod, at what point will the system (rod+ball) rotate instantaneously? Will it be the center of mass of the rod? Or the other end of the rod? Or the new center of mass of rod+ball system?

I had a similar question in which there were two identical balls A and B connected by a light rod instead of a rod (as in this case), and the velocities of A and B were asked immediately after collision (ball striking B). The answer given was vA=0 m/s and vB=non-zero.
Now this should mean that the system was rotating about A as vA=0 m/s.
Kindly explain.
 
Last edited:
cheapstrike said:
Hello,
I have a doubt related to this problem. When the ball strikes the rod, at what point will the system (rod+ball) rotate instantaneously? Will it be the center of mass of the rod? Or the other end of the rod? Or the new center of mass of rod+ball system?

I had a similar question in which there were two identical balls A and B connected by a light rod instead of a rod (as in this case), and the velocities of A and B were asked immediately after collision (ball striking B). The answer given was vA=0 m/s and vB=non-zero.
Now this should mean that the system was rotating about A as vA=0 m/s.
Kindly explain.
You should really post this as a separate problem. Point A can be instantaneously at rest if the speed of its rotation about the centre of mass equals the linear speed of the centre of mass. Then, at one point during each rotation it is instantaneously at rest.

This is similar to each point on the rim of rolling wheel being instantaneously at rest as it touches the ground.
 
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PeroK said:
You should really post this as a separate problem. Point A can be instantaneously at rest if the speed of its rotation about the centre of mass equals the linear speed of the centre of mass. Then, at one point during each rotation it is instantaneously at rest.

This is similar to each point on the rim of rolling wheel being instantaneously at rest as it touches the ground.
Ok.. so the rotation will be about center of mass but if the velocity of A w.r.t. C.M. = velocity of C.M. w.r.t. ground and both being in opposite directions, A will be at rest at that time frame.
Also, in the question asked above, will the location of point of rotation on rod change after ball strikes rod or will it be same as C.M. of rod. I guess it will be same.
Thanks, I'll keep in mind to make a separate thread for asking anything new. :D
 
cheapstrike said:
Ok.. so the rotation will be about center of mass but if the velocity of A w.r.t. C.M. = velocity of C.M. w.r.t. ground and both being in opposite directions, A will be at rest at that time frame.
Also, in the question asked above, will the location of point of rotation on rod change after ball strikes rod or will it be same as C.M. of rod. I guess it will be same.
Thanks, I'll keep in mind to make a separate thread for asking anything new. :D
The free motion of a rigid body can be decomposed into the motion of any point and a rotation about that point.

But, only the centre of mass moves in a straight line.

There was another thread about this recently, which you could search for. If you search my posts for rigid body.
 
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  • #11
Can someone please solve thus question... I am struggling
 
  • #12
Jitesh Kakkad said:
Can someone please solve thus question... I am struggling
As with the initial post on the thread, you need to show some attempt.
Better still, create a new thread and complete the template.
 
  • #13
image.jpg
image.jpg
 

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  • #14
My attempts
 
  • #15
Jitesh Kakkad said:
View attachment 218849 View attachment 218848
You seem to have assumed that the final velocity of the mass centre of the rod, V, is ωL/2. That would be true if the other end of the rod were to remain at rest initially, but it will not.
You need another equation. What law can you use?
 
  • #16
Hey... I just need you to tell me from which end does it rotate (together it translates too) ... from the end should I consider axis of rotation or should it be the centre of mass
 
  • #17
Jitesh Kakkad said:
Hey... I just need you to tell me from which end does it rotate (together it translates too) ... from the end should I consider axis of rotation or should it be the centre of mass
You cannot know in advance where the instantaneous centre of rotation is. Treat it as a translation plus a rotation. It does not matter which point you treat it as rotating about, so you might as well take it as a translation V of the mass centre plus a rotation ω about the mass centre.

The first three lines of your working were fine (except that in the second line the powers of 2 on the left hand side should not be there, but you seem to have realized and not carried them into the third line).
But in the fourth line you substitute V for Lω/2, which is not correct. You do not know the relationship between V and ω at this stage. Another equation is needed. What other conservation law can you use?
 
  • #18
Maybe linear momentum conservation
 
  • #19
Jitesh Kakkad said:
Maybe linear momentum conservation
Right.
 
  • #20
Isn't m=M in that case (which is wrong)
 
  • #21
Jitesh Kakkad said:
Isn't m=M in that case (which is wrong)
You got that result because in that attempt you ignored the energy that had gone into rotation.
 

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