Conservation of Momentum of a point particle

[eprparadox]

Homework Statement

A uniform stick of length L and mass M lies on a frictionless horizontal surface. A point particle of mass m approaches the stick with speed v on a straight line perpendicular to one end of the stick and intersects the stick at one end, as shown above (FIGURE ATTACHED). After the collision, which is elastic, the particle is at rest. The speed V of the center of mass of the stick after the collision is

a) (m/M)*v
b) [m/(m+M)]*v
c) Sqrt(m/M)*v
d) Sqrt([m/(m+M)])*v
e) (3m/M)*v

Homework Equations

KEi = KEf
Pi = Pf

The Attempt at a Solution

So I know the answer to this problem is A.

When solving this problem with conservation of momentum, I think you can imagine the ball hitting the center of mass with some speed v and then solve this is as we would any one-dimensional problem. Is that a correct way of thinking about this?

You would then get, mv = MV, which gives V = (m/M)*v.

My real question though is how can we solve it with energy considerations. I know the initial energy is .5*m*v^2 and this goes into rotating the stick and moving it translationally.

So we get .5*m*v^2 = .5*I*w^2 + .5*M*^2
(w = angular velocity of stick)


Can I not say in this instance (as we do for the pure rolling of a ball) that V= L*w? If you do that and then solve for V, you will not come back with (m/M)*v as the answer so it must be wrong, but I don't know why you can't say that.

Any explanations would be great. Thank you very much!

 

[tiny-tim]

hi eprparadox!

(have an omega: ω and a square-root: √ and try using the X² icon just above the Reply box )

… After the collision, which is elastic, the particle is at rest. The speed V of the center of mass of the stick after the collision is
…
When solving this problem with conservation of momentum, I think you can imagine the ball hitting the center of mass with some speed v and then solve this is as we would any one-dimensional problem. Is that a correct way of thinking about this?

yes, linear momentum is conserved (in the 3D case as well as the 1D case) completely separately from rotation

My real question though is how can we solve it with energy considerations. I know the initial energy is .5*m*v^2 and this goes into rotating the stick and moving it translationally.

So we get .5*m*v^2 = .5*I*w^2 + .5*M*^2
(w = angular velocity of stick)

Can I not say in this instance (as we do for the pure rolling of a ball) that V= L*w?

no, v = ωr only applies to rolling (without slipping) … r is the distance from the c.o.m. to the point of contact, and if there is rolling, the point of contact is instantaneously stationary, and therefore it is the centre of rotation, and so the c.o.m. must be moving with speed ωr

here, you have no idea where the centre of rotation is!

you can only find ω from the energy equation

 

[eprparadox]

Hey thank you very much for your response!