What Is the Distance Traveled by a Disc on a Block Before Coming to Rest?

[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)

 

[nlightner]

Your attempt at solving this problem is on the right track. However, there are a few mistakes in your calculations.

Firstly, when using the conservation of linear momentum equation, the mass of the block should be included in the calculation. The correct equation should be:

m1u1 = m1v1 + m2v2

0.2(6) = 0.2(0) + 1(1)

1.2 = 1

This means that the velocity of the block after the collision is 1 m/s, and the velocity of the disc is 0 m/s.

Secondly, when calculating the distance traveled by the disc on the block, you used the formula v^2-u^2=2as. However, this formula is only valid for objects with constant acceleration. In this case, the acceleration of the disc is not constant as it experiences friction. Instead, you should use the formula v^2 = u^2 + 2as, where v=0 m/s and u=6 m/s. This gives a distance of 18/12 m or 1.5 m.

Lastly, the time required for the disc to stop should be calculated using the equation v=u+at, where v=0 m/s and u=6 m/s. This gives a time of 6/6 s or 1 s.

Therefore, the disc travels a distance of 1.5 m on the block before coming to rest, and the time required for the disc to stop is 1 s. This is consistent with the distance and time calculated for the block, which is also 1.5 m and 1 s respectively.

In conclusion, the mistake in your calculation was not including the mass of the block in the conservation of linear momentum equation, and using the incorrect formula for calculating the distance traveled by the disc on the block. By correcting these mistakes, you should arrive at the correct answer of 1.5 m for the distance traveled by the disc on the block before coming to rest.