Conservation laws in mechanics

[chingcx]

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 traveled by the disc on the block before it comes to rest relative to the block.

Homework Equations

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]

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 traveled 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)