Angular Momentum Needed Variable Velocity

[AirChambz]

Homework Statement

A steel ball fastened to the end of a 1.5 m long string to form a pendulum is
released in the horizontal position. At the bottom of its swing, the ball collides elastically
with a block of equal mass resting on a frictionless surface. What is the speed of the block
just after the collision?

http://img686.imageshack.us/img686/3459/phys.png

R=1.5m
M_ball_=M_block_

Homework Equations

I=mr^2
L_A_=r*p*sin(theta)
L_A_=I(omega)

The Attempt at a Solution

I did not know where to start however I tried equating I(omega)=r*p*sin(theta) since momentum is conserved the momentum of the pendulum should turn into the momentum of the block, also since the masses are equal whatever final momentum you get should be the answer

Anyway the answer turns out to be 5.4 m/s in the end just am not able to get there...

 

[dr_k]

Homework Statement

A steel ball fastened to the end of a 1.5 m long string to form a pendulum is
released in the horizontal position. At the bottom of its swing, the ball collides elastically
with a block of equal mass resting on a frictionless surface. What is the speed of the block
just after the collision?

http://img686.imageshack.us/img686/3459/phys.png

R=1.5m
M_ball_=M_block_

Homework Equations

I=mr^2
L_A_=r*p*sin(theta)
L_A_=I(omega)

The Attempt at a Solution

I did not know where to start however I tried equating I(omega)=r*p*sin(theta) since momentum is conserved the momentum of the pendulum should turn into the momentum of the block, also since the masses are equal whatever final momentum you get should be the answer

Anyway the answer turns out to be 5.4 m/s in the end just am not able to get there...

Use energy conservation to determine the speed of the pendulum bob just before it hits the block. Now you know the intitial speed of the block before the collision.

Next draw a picture: (i) the bob and the block just before the collision, and (f) the bob and the block just after the collision:
(i) means initial
(f) means final

What is conserved?

 

[AirChambz]

E = W + Q Q=0 -> E=W W= Fd -> W= m*g*R?

 

[ideasrule]

Well yes, W=mgR, but you didn't need to involve Q; this isn't a thermodynamic problem.

If gravity does mgR of work, what's the speed of the bob at the bottom of its swing?

 

[AirChambz]

...sqrt(14.7)? (3.83m/s) with m canceled out

 

[dr_k]

E = W + Q Q=0 -> E=W W= Fd -> W= m*g*R?

This is a two-step problem.

First, determine the speed of the bob at the bottom of the swing, just before it hits the block. You can do this using conservation of energy. Work, of course, is related, but the power of energy conservation techniques lies in using path-independent potential energy functions, which are related to the work done by conservative forces.

Draw a picture (i) bob at the top of swing, and (f) bob at the bottom of swing, and apply

$$E_i= E_f$$
$$K_i+U_i = K_f+U_f$$

You'll only need the potential functions for gravity. Now you'll know the final speed of the bob, at the bottom of the swing. Now, start a new clock at (i) initial, where the bob is just about to hit the block, and then a (f) picture, after the block and bob have collided. Both the block and bob have final velocities.

What is conserved during the collision? The word "elastic" gives you a hint about one of the conserved quantities...it's in your text.