Minimum width of belt and minimum speed of the disc

Question

A horizontal conveyor belt is running at a constant speed $v_b$ = 3.0 m/s.
A small disc enters the belt moving horizontally with a velocity $v_0$ = 4.0
m/s that is perpendicular to the velocity of the belt. Coefficient of friction
between the disc and the belt is 0.50.

My try :

Comments

This 2nd part of this question is the same as [A block is pushed onto a conveyor belt](http://physics.qandaexchange.com/?qa=205/a-block-is-pushed-onto-a-conveyor-belt&show=205#q205).

Answer 1

(a) Your calculation is very good but you have not reached the end of it.

The distance of $2.5m$ which the disk travels across the belt is measured diagonally. It is the hypotenuse of a $3:4:5$ triangle. The perpendicular distance across the belt is $2.0m$. So that is the minimum width.

(b) This is the same problem as in A block is pushed onto a conveyor belt.

Initially the velocity of the disk over the ground is $OA=4m/s$ along the $Oy$ axis. When the disk has come to rest relative to the conveyor belt its velocity over the ground is $OB=3m/s$ along the $Ox$ axis. At intermediate times the velocity of the disk over the ground is represented by points on the line $AB$.

At point $C$ the magnitude $OC$ of the velocity over the ground is minimum, where $OC\perp AB$. $OAB$ and $COB$ are similar triangles. Therefore $OC=\frac45 \times 3=2.4m/s$.