# Conservation laws in mechanics

## Homework Statement

A 0.2 kg disc slides down from smooth track of height 1.8m. It arrives at a rough 1 kg block resting on a smooth surface. The friction between them is 1.2N. Find the distance travelled by the disc on the block before it comes to rest relative to the block.

## The Attempt at a Solution

use observer frame:
velocity=6 m/s
then I use conservation of linear momentum, because only internal forces.
m1u1=m1v1+m2v2
0.2(6)=1.2(v)
v=1 m/s

f=1.2N to right on the block, to left on the disc
a of disc=-6m/s^2
a of block= 1 m/s^2

for the disc, v^2-u^2=2as
-35=2(-6)s
s=35/12 m

for the block,
1=2(1)(s)
s=0.5 m

so it moves 2.41 m on the surface

but when I find the time required
for the disc, v=u+at
1=6+(-6)t
t=5/6 s

for the block
1=0+(1)(t)
t= 1s

They are different?

I know my answer of 2.41 m must be wrong, but where is my mistakes?
thank you!

tiny-tim
Homework Helper
Hi chingcx! A 0.2 kg disc slides down from smooth track of height 1.8m. It arrives at a rough 1 kg block resting on a smooth surface. The friction between them is 1.2N. Find the distance travelled by the disc on the block before it comes to rest relative to the block.

then I use conservation of linear momentum, because only internal forces.
m1u1=m1v1+m2v2
0.2(6)=1.2(v)
v=1 m/s

I don't follow this: it's not an impulsive collision f=1.2N to right on the block, to left on the disc
a of disc=-6m/s^2
a of block= 1 m/s^2
Correct method, but check your figures.

for the disc, v^2-u^2=2as …

I'm not sure what you're doing here.

You can either use v2 = u2 + 2as for each body, and find what s gives the same v,

or (especially since it seems they want you to use conservation) you can use change in energy = work done (1/2 m v2 = 1/2 m u2 + Fs), which amounts to the same thing (and they give you F anyway, plus this avoids working out what a is) 