2 simple Machanics Qs plz~~~ (1 Viewer)

[Hikari Clover]

A particle of mass m moves in a horizontal straight line away from a fixed point O in the line. The particle is resisted by a force mkv<SUP>3/2</SUP>, where k is a positive constant and v is the speed. When t=0, v=u>0. Show that the particle is never brought to rest and that its distance from O is at most (2/k) [FONT=宋体]√[/FONT]u.

 

[ssglain]

Q1.
ma = -mkv<SUP>3/2</SUP>
a = dv/dt = -kv<SUP>3/2 </SUP>=> dt/dv = -(1/k)v<SUP>-3/2</SUP>
Integrate to find t = (2/k)v<SUP>-1/2</SUP> + C
Clearly, v cannot = 0 -> the particle is never brought to rest.
It might make your teacher/marker happier if you use the condition (t = 0, v = u) to find C before drawing that conclusion - although it is irrelevant.
C = (2/k)u<SUP>-1/2 </SUP>=> t = (2/k)v<SUP>-1/2</SUP> + (2/k)u<SUP>-1/2</SUP>

Also, a = v dv/dx = -kv<SUP>3/2 </SUP>=> dx/dv = (-1/k)v<SUP>-1/2 </SUP>
Integrate to find x = -(2/k)v<SUP>1/2</SUP> + C
Use the condition (x = 0, v = u) to find C = (2/k)u<SUP>1/2</SUP>
Then x = -(2/k)v<SUP>1/2</SUP> + (2/k)u<SUP>1/2</SUP>
The max distance is reached when v = 0 (or rather, v -> 0 in this case).
Therefore, max x = (2/k)u<SUP>1/2</SUP>

Can't help with Q2 - it's incomplete.

 

[Hikari Clover]

thx

Q2 is here