A particle of mass approaches a region of force starting from r = +∞. The potential energy function in terms of distance from the origin is given by... best 1

A particle of mass approaches a region of force starting from r = +∞. The potential energy function in terms of distance from the origin is given by…

Question :

A particle of mass approaches a region of force starting from r = +∞. The potential energy function in terms of distance from the origin is given by,

$U_{(r)}={K}/{2a^3}{(3a^2-r^2)}$ for 0 ≤ r ≤ a
$U_{(r)}={K}/{r}$ for r ≥ a where K>0 is a positive constant.

(a) Derive the force F_(r) and determine whether it is repulsive or attractive.

(b) With what velocity should the particle start at r = ∞ to cross over to other side of the origin.

Solution:

(a) F_(r) = ? Nature of force whether it is repulsive or attractive =?

Concept: F= - {dU}/{dr}

$U_{(r)}={K}/{2a^3}{(3a^2-r^2)}$
Differentiating above equation wrt dr
For 0 ≤ r ≤ a $F= - {dU}/{dr} = - {d}/{dr} [{K}/{2a^3}{(3a^2-r^2)}]$

$F = - {d}/{dr} [{K}/{2a^3}{(0 - 2r)}]$
$F = [{K r}/{a^3}]$ F is positive.

For r ≥ a
$U_{(r)}={K}/{r }$
Differentiating above equation wrt dr
F= - {dU}/{dr} = - {d}/{dr} [{K}/{r}]
$F = {K}/{r^2}$ F is positive.

From above it is clear that F is positive in both cases so Force is repulsive in nature.

(b) at r = ∞ v = ? just to cross over to the other side of the origin.

From Law of conservation of energy,

(ME)₁ = (ME)₂

(ME)₁ = KE + PE

$KE = {1/2}{mv^2}$

PE =?
For r ≥ a $U_{(r)}={K}/{r }$
so, at r = ∞ U = 0

$ME_1 = KE + PE = {1/2}{mv^2}{+0} = {1/2}{mv^2}$

(ME)₂ = KE + PE

KE = 0 ( as velocity of particle should be equal to zero, just to pass other side of origin )

At origin PE = ?
For 0 ≤ r ≤ a $U_{(r)}={K}/{2a^3}{(3a^2-r^2)}$

at origin r = 0
$PE={K}/{2a^3}{(3a^2-0)} ={K}/{2a^3}{(3a^2)} ={3K}/{2a}$

$so, ME_2= {0+} {3K}/{2a}={3K}/{2a}$

(ME)_1 = (ME)_2

$so, {1/2}{mv^2} = {3K}/{2a}$

v = sqrt{(3K)/am}

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