# Conservation of momentum and energy

#### royguitarboy

1. Homework Statement

A pendulum consists of a 0.5 kg bob attached to a string of length 1.6 m. A block of mass m rests on a horizontal frictionless surface. The pendulum is released from rest at an angle of 53° with the vertical and the bob collides elastically with the block. Following the collision, the maximum angle of the pendulum with the vertical is 5.73°. Determine the two possible values of the mass m.

2. Homework Equations

Conservation of momentum and energy

3. The Attempt at a Solution

I got 2.16 kg for one...which isn't right.

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#### Mentz114

Gold Member
Hi, you must show how you got that result, so we can fix it if it's wrong.

#### royguitarboy

Ki(objec1t)+Ki(object2)+Ui(object1)+Ui(object2)=Kf(object1)+Kf(object2)+Uf(object1)+Uf(object2)

K= kinetic energy------U= potential energy-------i=initial------f=final
K=1/2mv^2-----------U=mgh

0+0+0+cos(53)*1.6*.5*9.8=sin(5.73)(1.6)+(.5)(cos(5.73)*1.6)+m2vf^2

Change in kinetic energy = 0 so,

(m1v1i^2+m2vi^2)+(m1v1f^2+m2v2f^2)=0 algebra...

m1(v1f^2-v1i^2)=m2(v2i^2-v2f^2) where m2=3.76/v^2 from the conversation of momentum part

then substituted

v^2=.8/3.76

#### Mentz114

Gold Member
0+0+0+cos(53)*1.6*.5*9.8=sin(5.73)(1.6)+(.5)(cos(5 .73)*1.6)+m2vf^2
OK, the LHS is the potential energy at the start. That should be sin(53) ( easy to remember because as the angle increases so does the pot.)
On the right we should have the final potential energy of the bob plus the kinetic E of the mass.

sin(5.73)*1.6*.5*9.8 + m2v2f^2
To get the momentum equation, work out the velocity of the bob at the lowest point, and you've got it ( after a bit of algebra)

#### royguitarboy

Sorry I'm a little confused. I'm not sure how to get the velocity at the lowest point other than it being sin(0)*1.6 which would be 0. And doesn't " sin(5.73)*1.6*.5*9.8 + m2v2f^2" go into the momentum equation?

If it's easier to talk on aim, my sn is:

Royguitarboy246

Edit: Wouldn't that be cos(0)*1.6 which would then equal 1.6?

Last edited:

#### Mentz114

Gold Member
You get the velocity at the lowest point by equating the initial pot energy of the bob with its KE at the lowest point.

sin(57)*1.6*0.5*9.8 = 0.5*0.5*vf^2

You can get the velocity of the bob after the collision using the same approach.

And doesn't " sin(5.73)*1.6*.5*9.8 + m2v2f^2" go into the momentum equation?
No, that's energy ! The momentum equation is

0.5*( vi-vf) = m*v2 ( v2 is the speed of the mass after the collision, vi, vf are bob speeds before and after.)

I have to go now. You can do it. Use a pencil, paper and eraser, and list your quantities carefully.

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