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cyrus_sofia · Aug 1, 2010, 10:13 AM

A pendulum with a bob of mass 30 grams is lifted to a height of 5.00 cm above its lowest point and allowed to fall so that it collides elastically with a block of mass 0.0 gramss that rests on a smooth horizontal floor.The block stop 5.00 seconds after the collision. Assuming no sound is generated during the colliion. Determine the :
a) speed of the bob just before it collides with the block
b) distance travelled by the block before it stop
c) decelleration of the block

Unknown008 · Aug 2, 2010, 12:52 AM

1. Using the principle of conservation of energy, you know that the potential energy gained by the bob upon raising it is the same as the kinetic energy it has before hitting the wooden block.

Then, you got:

$E_p = E_k$

$mgh = \frac12 mv^2$

$gh = \frac2 v^2$

You know g, the acceleration due to gravity, you know the height it was raised (in metres). Find the velocity of the bob.

2. Assuming that energy is conserved after collision, you can find the velocity transferred to the block fro the bob using the kinetic energy formula.

Using the formulae of kinematics: $s = ut + \frac12 at^2$ and $v^2 = u^2 + 2as$ ,

you get:

$s = ut + \frac12 (\frac{v^2 - u^2}{2s})t^2$

Here, u = initial velocity of block
v = final velocity of block (that is 0)
t = duration of motion
s = distance moved by block

3. Now that you have the distance, use one of the equations of kinematics above to get the decceleration.