# Is momentum conserved in the collision of a ball with a hanging rod?

Suppose we have a situation like A ball of some mass $m$ with some velocity collides with rod hinged at point $A$. Is momentum conserved in this situation? I know that hinge will give impulsive force on rod but that is internal force when we take ball and rod together as a system so that shouldn't stop us using conservation of momentum equation

• At the instant of collision, even in a gravitational field, I believe conservation of angular momentum will hold. Certainly, gravity will start affecting things immediately, but at the instant of collision, I feel confident conservation of angular momentum can be applied. – Inquisitive Feb 10 '15 at 23:47
• @Sigma Your accepted answer to this question is wrong.It will misguide future viewers. – user74370 Sep 27 '16 at 0:02

Generally speaking linear momentum cannot be conserved in this case as hinge will exert a (external) force on rod at all but one position.

Note:Angular momentum is definitely conserved about hinged point since the force applied by hinge produces zero torque here.

We may want to find that one position for which linear momentum conservation holds.We proceed as follows:

$mv_od=mv_{final}d+(M\dfrac{l^2}{3})\omega$ -----(1)

$mu=mv_{final}+M(\dfrac{l}{2}\omega)$-------------(2) [holds since force by hinge is zero for this position $d$]

Solving we get $d=\dfrac{2l}{3}$

In this cases, momentum is not conserved because of the action of gravity as an external force. When you have a pivoted rod, as in your problem, you can use basically two conservation laws:

a) conservation of energy, if the collision is assumed as perfectly elastic;

b) conservation of angular momentum about the pivot.

As regards b), indeed, if we choose the pivot point A as the point in which to calculate torque, then the torque about the pivot is

$$\tau_A^{tot}=\vec r_{A,A}\times\vec F_{pivot} + \vec r_{A,P}\times\vec F_{ball,rod}+\vec r_{A,P}\times\vec F_{rod,ball}+\vec r_{S,cm}\times\vec F_{g,rod}$$

where:

• $\vec F_{m,r}$ is the force exerted by the ball on the rod, and $\times\vec F_{r,m}$ is the force exerted by the rod on the ball. As you noticed, these two cancel because of Newton's third law, so $\times\vec F_{ball,rod}=-\times\vec F_{rod,ball}$ and they do not give any net torque;
• If the collision is istantaneous, then the gravitational force is parallel to the position vector of the center of mass $\vec r_{A,cm}$, therefore $\vec r_{A,cm}\times\vec F_{g,rod}=\vec 0$.

• $r_{A,A}$ is the position vector of $A$ with respect to $A$, so it is identically zero by definition.

In the end, we have showed that the total torque $\vec\tau^{tot}_{A}=0$, and using the second cardinal equation (i.e. Newton's 2nd Law applied to a material mechanical system of points), the total angular momentum is conserved.

• This answer is wrong.Inconsistency of linear momentum is due to reaction force by hinge and not due to gravity.At the moment of collision gravity exerts no impulse. – user74370 Sep 27 '16 at 0:01

I think that the law of conservation will hold good in this situation because there is no external force acting on the system. Because like you said the impulsive force by the hinge is internal and no other force is a acting on the system.

• Alex, what is gravity? – Ryan Unger Feb 10 '15 at 20:55