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## 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!