Query 10

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course MTH 279

Query 10 Differential Equations*********************************************

Question: 3.7.4. Solve the equation m dv/dt = - k v / (1 + x), where x is position as a function of t and v is velocity as a function of t.

How far does the object travel before coming to rest?

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Your solution:

dv/dt = dv/dx*dx/dt = v*dv/dx

mv*dv/dx = -kv/(1+x)

dv/dx = -k/(m(1+x))

ʃdv = ʃ-k/(m(1+x))dx

v = -k/m*ln(1+x) + C

v_0 = C

v(x) = -k/m*ln(1+x)+v_0

0 = -k/m*ln(1+x) + v_0

x = ±e^(v_0*m/k)-1

The object will travel e^(v_0*m/k)-1 or -e^(v_0*m/k)-1before coming to rest.

@&

Good.

The integral leads to a multiply of ln | 1 + x | so from that point we have

| 1 + x | = e^(m / k * v_0)

If x > -1 then | 1 + x | = 1 + x and we get

x = e^(m / k * v_0) - 1.

If x < -1 then | 1 + x | = -1 - x and we get

x = -e^(m / k * v_0) - 1.

*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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Question:

3.7.6. A vertical projectile has initial velocity v_0, and experiences drag force k v^2 in the direction opposite its motion. Assume that acceleration g of gravity is constant. Find the maximum height to which the projectile rises.

If the initial velocity is 80 m/s and the projectile, whose mass is .12 grams, rises to a height of 40 meters (estimated quantities for a plastic BB), what is the value of k?

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Your solution:

dv/dt = dv/dx*dx/dt = v*dv/dx

mvdv/dx = -kv^2-mg

vdv/dx = -kv^2/m - g

ʃvdv/(k/m*v^2+g) = ʃ-dx

m/(2k)*ln(kv^2/m+g) = -x + C

v = sqrt([m*Ae^(-2kx/m)-mg]/k)

v_0 = sqrt((mA-mg)/k)

A = kv_0^2/m + g

v(x) = sqrt([m(kv_0^2/m+g)e^(-2kx/m)-mg]/k)

x_max : v(x) = 0

x_max = mln((kv_0^2+gm)/(gm))/(2k)

@&

My expression at this point was

x_max = m / (2 k) * ln ((k / (m g) * v_0^2 + 1) )

which agrees with yours.

*@

40 = 0.12ln((80^2*k + 9.81*0.12)/(0.12*9.81))/(2k)

k ≈ 0.00501 g/m

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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Question:

3.7.8. Mass m accelerates from velocity v_1 to velocity v_2, while constant power is exerted by the net force. At any instant power = force * velocity (physics explanation: power = dw / dt, the rate at which work is being done; w = F * dx so dw/dt = F * dx/dt = F * v).

How far does the mass travel as it accelerates?

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Your solution:

F = P/v

F = m dv/dt

dv/dt = dv/dx * dx/dt = v dv/dx

mv dv/dx = P/v

ʃmv^2 dv = ʃPdx

mv^3/3 = Px + C

mv_2^3/3 - mv_1^3/3 = P(x_2-x_1)

d = x_2-x_1

d = mv_2^3-mv_1^3/(3P)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

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Question:

3.7.11. An object falls to the surface of the Earth from a great height h, and as it falls experiences a drag force proportional to the square of its velocity. Assume that the gravitational force is - G M m / r^2.

What will be its impact velocity?

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Your solution:

F = -GmM/r^2

mdv/dt = -GmM/r^2 + kv^2

mvdv/dr = -GmM/r^2 + kv^2

dv/dr = -GM/(r^2*v) + kv/m

dv/dr - kv/m = -GM/r^2 * v^(-1)

v dv/dr - kv^2/m = -GM/r^2

u = v^2 ; u’ = 2vv’

u’/2 - ku/m = -GM/r^2

u’ - 2ku/m = -2GM/r^2

µ = e^(ʃ2k/mdr) = e^(2kr/m)

ʃd/dr(e^(2kr/m)u)dr = ʃ-2GM/r^2*e^(2kr/m) dr

e^(2kr/m)*u = -2GM ʃe^(2kr/m)/(r^2)dr

u = -2GMe^(-2kr/m)*ʃe^(2kr/m)/r^2 dr

v = sqrt(-2GMe^(-2kr/m)*ʃe^(2kr/m)/r^2 dr)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I think this correct, or very nearly. I may have made an algebraic mistake at some point or something of that matter.. but hopefully not.

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Self-critique rating: 2"

&#This looks good. See my notes. Let me know if you have any questions. &#