Elastic Collision between a Ball and a Stick

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Homework Statement


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A mass ##m## travels in a straight line with velocity ##v_0## perpendicular to a uniform stick of mass ##m## and length ##l##, which is initially at rest.. The distance from the center of mass of the stick to the path of the travelling mass is ##h##. Now the travelling mass ##m## collides elastically with the stick, and the centre of the stick and mass ##m## are observed to move with equal speed ##v## after the collision. Assuming the travelling mass can be trated as a point mass, and the moment of inertia of the stick about its centre is ##I=\frac {ml^2}{12}##, calculate the distance ##h##.

Homework Equations


For elastic collisions,
I. Kinetic Energy is conserved.
II. Linear Momentum is conserved.
Additionally, since there is no external torque about the centre of mass
III. Angular Momentum about the centre of mass is conserved.
$$((E_k)_{ball})_i={\frac 1 2}m{v_0}^2$$
$$((E_k)_{ball})_f={\frac 1 2}mv^2$$
$$((E_k)_{stick})_i=0$$
$$((E_k)_{stick})_f={\frac 1 2}mv^2 + {\frac 1 2}I\omega^2$$

$$((\vec p_{ball})_i=mv_0 \hat i$$
$$((\vec p_{ball})_f=mv \hat i$$
$$((\vec p_{stick})_i=0$$
$$((\vec p_{stick})_f=mv(\hat i cos \theta + \hat j sin \theta)$$

$$((\vec L_{ball})_i=\vec{r} \times m\vec{v_0}$$
$$((\vec L_{ball})_f=0$$
$$((\vec L_{stick})_i=0$$
$$((\vec L_{stick})_f=I\vec \omega + ?$$


The Attempt at a Solution


Conservation of Linear Momentum gives
$$\sum {\vec p}_i=\sum {\vec p}_f$$
$$\Rightarrow mv_0 \hat i=mv\hat i + mv(\hat i cos \theta + \hat j sin \theta)$$
Equating coefficients of ##\hat i## and ##\hat j## -
$$v sin \theta =0$$
$$v_0=v(1+cos \theta)$$
Taking square and adding and simplifying
$$2v=v_0$$

From Conservation of Kinetic Energy
$$E_i = E_f$$
$$\Rightarrow {\frac 1 2}m{v_0}^2={\frac 1 2}mv^2+{\frac 1 2}mv^2 + {\frac 1 2}I\omega^2$$
After simplification, this gives
$$m({v_0}^2 - v^2)=I{\omega}^2$$

Using $$2v=v_0$$, this gives
$$I{\omega}^2=2mv^2$$

But I cannot calculate the angular momentum of the stick.
Is it $$((\vec L_{stick})_f=I\vec \omega + m\vec v ({\frac l 2}-r)$$

If someone has any suggestion, it might be helpful.
 

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Answers and Replies

  • #2
vela
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Think about what origin you are using to calculate the angular momentum of the system. The stick will conceivably have angular momentum due to its rotation about its center of mass and also due to the motion of its center of mass. The ball, likewise, may have angular momentum after the collision because it's still moving.

One mistake I noticed was that angular momentum is ##\vec{r}\times\vec{p}##. You have the two vectors reversed in your expression for the angular momentum of the ball.
 
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One mistake I noticed was that angular momentum is ##\vec{r}\times\vec{p}##. You have the two vectors reversed in your expression for the angular momentum of the ball.
Thanks for pointing out the typing error @vela !

Think about what origin you are using to calculate the angular momentum of the system. The stick will conceivably have angular momentum due to its rotation about its center of mass and also due to the motion of its center of mass. The ball, likewise, may have angular momentum after the collision because it's still moving.
If the ball has angular momentum after the collision that will complicate the problem, even more and don't know how to calculate that.
For the stick, the angular momentum due to the rotational motion about its centre of mass is ##I{\omega}^2##, but since the ball strikes at ##h## distance above the centre of mass of the stick, I'm not sure what is the angular momentum due to the motion of its centre of mass.
Where is the opposite torque acts? At the lower end of the stick?
 
  • #4
haruspex
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If the ball has angular momentum after the collision that will complicate the problem, even more and don't know how to calculate that.
For the stick, the angular momentum due to the rotational motion about its centre of mass is ##I{\omega}^2##, but since the ball strikes at ##h## distance above the centre of mass of the stick, I'm not sure what is the angular momentum due to the motion of its centre of mass.
Where is the opposite torque acts? At the lower end of the stick?
You have to be a bit careful about choosing reference axes for dynamics problems. The safest is to choose a fixed point in space. In this case, the initial position of the stick's centre would be simplest.
Both the initial and final angular momenta of the ball can be computed by application of ##\vec L=\vec r\times\vec p##.
 
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  • #5
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You have to be a bit careful about choosing reference axes for dynamics problems. The safest is to choose a fixed point in space. In this case, the initial position of the stick's centre would be simplest.
Both the initial and final angular momenta of the ball can be computed by application of ##\vec L=\vec r\times\vec p##.
I understand that.
What I'm having difficulty in understanding is that since the ball strikes the stick at ##h## off the centre of the stick, where will the equal and opposite torque work? Is it ##\frac l 2## or ##h##?

From the conservation of linear momentum, I obtained $$v_0=2v$$
From the conservation of total kinetic energy, I obtained $$I{\omega}^2=m({v_0}^2-2v^2)$$
Which gives, $$I{\omega}^2=m[(2v)^2-2v^2]$$
Substitution of the moment of inertia of the stick $$I={\frac {1} {12}}ml^2$$ gives
$${\frac {1} {12}}ml^2{\omega}^2=2mv^2$$
$$\Rightarrow \omega=2{\sqrt 6} {\frac v l}$$

So, if the opposite torque acts at ##h##, the angular momentum conservation gives $$\sum {\vec L_i}=\sum {\vec L_f}$$
$$\Rightarrow mv_0 h=mvh+I\omega$$
$$\Rightarrow m(2v) h=mvh+{\frac {1} {12}}ml^2{\sqrt 6} {\frac v l}$$
$$\Rightarrow mvh={\frac {1} {\sqrt 6}}mvl$$
Which gives $$h={\frac {1} {\sqrt 6}}l$$

Which is the desired answer.


But I'm still not really clear about the idea of the distance where the opposite torque work.

Say, I've a stick of length ##L##. If a force acts at one end of the stick i.e., at a distance ##\frac L 2## from the centre of mass of the stick then the torque acting is $$\vec N=\vec r \times \vec F$$ and the equal and opposite torque ##-\vec N## acts the other end at ##\frac L 2## from the centre of mass of the stick.
But what if the force acts at ##a## from the centre of mass of the stick?
Then the torque at ##a## is ##N=mva##.
Does the opposite torque, too, acts at ##a##, at the other side of the centre of mass of the stick?
Is it so?

Thanks @haruspex and @vela for your important suggestions. :smile::smile::smile:
 
  • #6
haruspex
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But I'm still not really clear about the idea of the distance where the opposite torque work.
You do not need to care. The equal and opposite torques are internal to the system.
But if you want to break it into ball and stick as separate systems then...
the equal and opposite torque acts the other end
... the equal and opposite impulses act at the point of impact. Not sure it means anything to ask where torques act. The torque from a force depends on your choice of reference axis, and to compare them you would necessarily use the same axis for both.
 
  • #7
Abhishek kumar
Thanks for pointing out the typing error @vela !If the ball has angular momentum after the collision that will complicate the problem, even more and don't know how to calculate that.
For the stick, the angular momentum due to the rotational motion about its centre of mass is ##I{\omega}^2##, but since the ball strikes at ##h# distance above the centre of mass of the stick, I'm not sure what is the angular momentum due to the motion of its centre of mass.
Where is the opposite torque acts? At the lower end of the stick?
if you choose centre of mass
as reference point say c which is mid point of stick initially at rest and apply the conservation of angular momentum about that piont you notice just after the collision centre of mass shifted some distance r and you can write angular momentum about point c
L(final)=Lcm+mr×v+mvh
But here r is parallel to v so this will be zero
 
  • #8
vela
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But I'm still not really clear about the idea of the distance where the opposite torque work.

Say, I've a stick of length ##L##. If a force acts at one end of the stick i.e., at a distance ##\frac L 2## from the centre of mass of the stick then the torque acting is $$\vec N=\vec r \times \vec F$$ and the equal and opposite torque ##-\vec N## acts the other end at ##\frac L 2## from the centre of mass of the stick.
In addition to what haruspex said, there is no equal and opposite torque on the stick. If there were, the net torque on the stick would be 0, and the stick would not end up rotating about its center of mass.

Newton's third law implies that when the ball exerts a force on the stick, the stick exerts an equal and opposite force on the ball. Because the forces act on different objects, the corresponding torques are on different objects. The torque on the stick gets it rotating after the collision while the torque on the ball just slows it down.
 

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