Assignment 23

course PHY201

That was a tricky homework assignment.

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Physics I

10-15-2007

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18:59:08

Query gen phy 7.27 bumper cars 450 kg at 4.5 m/s, 550 kg at 3.7 m/s, collision from back, elastic

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RESPONSE -->

For the problem we must find the velocities after the collision and then the change in momentum of both cars before and after collision.

For momentum:

m1 * v1 + m2 * v2(before) = m1 * v1 + m2 * v2(after)

(450kg)(4.5m/s) + (550kg)(3.7m/s) = 4060 kg * m/s

After Colliosion:

(v2 - v1) = (3.7m/s - 4.5m/s) = -.8m/s

v2 = v1 - .8

v2 = 4.5m/s - .8m/s = 4.42m/s

v1 = v2 - .8m/s

v1 = 3.7m/s - .8m/s = 3.62m/s

m1 * v1 +m2 * v2 =

(450kg)(3.62m/s) + (550kg)(4.42m/s) = 4060kg * m/s

4060kg * m/s(before) = 4060kg * m/s(after)

Velocities after the collision:

Car 1 = 3.62m/s

Car 2 = 4.42m/s

Now to find the momentum change.

Car 1 = m1 * v1(after) - m1 * v1(before)

(450kg)(3.62m/s) - (450kg)(4.5m/s) = -396kg * m/s

Car 2 = m2 * v2(after) - m2 * v2(before)

(550kg)(4.42m/s) - (550kg)(3.7m/s) = 396kg * m/s

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18:59:31

** For an elastic collision we have m1 v1 + m2 v2 = m1 v1' + m2 v2' and v2 - v1 = -( v2' - v1').

We substitute m1, v1, m2 and v2 to obtain

450 kg * 4.5 m/s + 550 kg * 3.7 m/s = 450 kg * v1 ' + 550 kg * v2 ', or

4060 kg m/s = 450 kg * v1 ' + 550 kg * v2 ' . Dividing by 10 and by kg we have

406 m/s = 45 v1 ' + 55 v2 '.

We also obtain

3.7 m/s - 4.5 m/s = -(v2 ' - v1 ' ) or

v1 ' = v2 ' - .8 m/s.

Substituting this into 406 m/s = 45 v1 ' + 55 v2 ' we obtain

406 m/s = 45 ( v2 ' - .8 m/s) + 55 v2 ' . We easily solve for v2 ' obtaining

v2 ' = 4.42 m/s. This gives us

v1 ' = 4.42 m/s - .8 m/s = 3.62 m/s.

Checking to be sure that momentum is conserved we see that the after-collision momentum is

pAfter = 450 kg * 3.62 m/s + 550 kg * 4.42 m/s = 4060 m/s.

The momentum change of the first car is m1 v1 ' - m1 v1 = 450 kg * 3.62 m/s - 450 kg * 4.5 m/s = - 396 kg m/s.

The momentum change of the second car is m2 v2 ' - m2 v2 = 550 kg * 4.42 m/s - 550 kg * 3.7 m/s = + 396 kg m/s.

Momentum changes are equal and opposite.

NOTE ON SOLVING 406 m/s = 45 ( v2 ' - .8 m/s) + 55 v2 ' FOR v2 ':

Starting with

406 m/s = 45 ( v2 ' - .8 m/s) + 55 v2 ' use the Distributive Law to get

406 m/s = 45 v2 ' - 36 m/s + 55 v2 ' then collect the v2 ' terms to get

406 m/s = -36 m/s + 100 v2 '. Add 36 m/s to both sides:

442 m/s = 100 v2 ' so that

v2 ' = 442 m/s / 100 = 4.42 m/s. *

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RESPONSE -->

Thank goodness this question was in the book. The program didn't state the entire question.

The program's statements are generally abbreviated and intended just as guideposts; the book's statement is the one that counts.

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20:59:01

Univ. 3.48. (not in 11th edition) ball 60 deg wall 18 m away strikes 8 m higher than thrown. What are the Init speed of the ball and the magnitude and angle of the velocity at impact?

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RESPONSE -->

Given the known values of dsy, dsx, and the angle of projection, I will calculate for Intial Velocity and Final Velocity with angle.

dsy = 8m

dsx = 18m

ay = -9.8m/s

ax = 0

Voy = Vo * sin(60 deg) = .5(Vo)

Vox = Vo * cos(60 deg) = .866(Vo)

Horizontal:

Vfx = Vox

dsx = Vox(dt)

18m = .5(Vo)(dt)

dt = 36 / Vo

Vo = 36 / dt

Vertical:

dsy = Voy(dt) - .5(g)(dt^2)

8m = (.866)(Vo)(36 / Vo) - (.5)(9.8m/s)[(36 / Vo)^2]

8m = 31.176 - (4.9)(1296 / Vo^2)

-23.176 = - 6350.4 / Vo^2

Vo^2 = 6350.4 / 23.176 = 274

Vo = sqrt(274) = 16.55m/s

Intial Velocity = 16.55m/s

Now I will solve for the magnitude and angle that the ball strikes the wall.

Voy = 16.55 * sin(60 deg) = 14.33m/s

Vox = 16.55 * cos(60 deg) = 8.275m/s

dt = dsx / Vox

dt = 18m / 8.275m/s

dt = 2.175s

Vfx = Vox

8.275m/s = 8.275m/s

Vfy^2 = Voy^2 - 2g(dsy)

Vfy^2 = (14.33m/s)^2 - 2(9.8m/s^2)(8m)

Vfy^2 = 205.35m^2/s^2 - 156.8m^2/s^2

Vfy = sqrt(48.55m^2/s^2) = 6.97m/s

Vfy = 6.97m/s

Now that Vfx and Vfy are solved for, I can calculate for Vfz.

Vfz = sqrt(Vfx^2 + Vfy^2)

Vfz = sqrt(8.275^2 + 6.97^2)

Vfz = sqrt(117.0565)

Vfz = 10.82m/s

tan-1(y/x)

tan-1(6.97 / 8.275) = 40 deg

The magnitude of the ball when it hits the wall:

10.82m/s @ -40deg

or

10.82m/s @ 320 deg

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20:59:56

** We know the following:

For y motion `dsy = + 8 m, ay = -g = - 9.8 m/s^2 and v0y = v0 sin(60 deg) = .5 v0.

For x motion `dsx = 18 m, ax = 0 and v0x = v0 cos(60 deg) = .87 v0, approx.

Assuming a coordinate system where motion starts at the origin:

The equation of motion in the x direction is thus

x = .5 v0 * t

and the equation of y motion is

y = .87 v0 t - .5 g t^2.

We know x and y at impact and we know g so we could solve these two equations simultaneously for v0 and t.

We begin by eliminating t from the two equations:

x = .5 v0 * t so

t = 2 x / v0.

Substituting this expression for t in the second equation we obtain

y = .87 v0 * (2 x / v0) - .5 g ( 2 x / v0) ^ 2. Multiplying both sides by v0^2 we obtain

v0^2 y = .87 v0^2 ( 2 x) - .5 g * 4 x^2. Bringing all the v0 terms to the left-hand side we have

v0^2 y - 1.73 v0^2 x = -2 g * x^2. Factoring v0 we have

v0^2 ( y - 1.73 x) = -2 g x^2 so that

v0 = +-sqrt(-2 g x^2 / ( y - 1.73 x) ) ) . Since we know that y = 8 m when x = 18 m we obtain

= +- sqrt( -9.8 m/s^2 * (18 m)^2 / ( 8 m - 1.73 * 18 m) ) = +-sqrt(277 m^2 / s^2) = +-16.7 m/s, approx..

We choose the positive value of v0, since the negative value would have the projectile moving 'backward' from its starting point.

Substituting this value into t = 2 x / v0 and recalling that our solution applies to the instant of impact when x = 16 m and y = 8 m we obtain

t = 2 * 18 m / (16.7 m/s) = 2.16 s.

Alternatively we can solve the system for v0 less symbolically and perhaps gain different insight into the meaning of the solution. Starting with the equations

x = .5 v0 * t and y = .87 v0 t - .5 g t^2

we see that impact occurs when x = .5 v0 t = 18 m so that t = 18 m / (.5 v0) = 36 m / v0.

At this instant of impact y = 8 m so substituting this and the t just obtained into the equation of motion for y we get

y = .87 v0 (36 m / v0 ) - .5 g (36 m / v0 )^2 = 8 m.

The equation .87 v0 (36 m / v0 ) - .5 g (36 m / v0 )^2 = 8 m is easily solved for v0, obtaining v0 = 16.7 m/s.

With this initial velocity we again confirm that t = 2.16 sec at impact.

Note that at t = 2.16 sec we get y = 14.4 m/s * 2.16 s - 4.9 m/s^2 * (2.16 s)^2 = 8 m, within roundoff error, confirming this solution.

We need the magnitude and direction of the velocity at impact. We therefore need the x and y components of the velocity at the t = 2.16 sec instant.

At this instant we have x and y velocities

vx = dx/dt = .5 v0 = 8.35 m/s and

vy = dy/dt = 14.4 m/s - 9.8 m/s^2 * t = 14.4 m/s - 9.8 m/s^2 * 2.16 s = -5.6 m/s, approx.

The velocity at impact therefore has magnitude sqrt( (8.25 m/s)^2 + (-5.6 m/s)^2 ) = 10 m/s, approx. and the angle is arctan(vy/vx) = arctan(-5.6 / 8.25) = -34 deg, approx.

At impact the object is moving at 10 m/s and at 34 deg below horizontal (i.e., it's on its way back down). **

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RESPONSE -->

This question was hard! I was lost at first, but I finally figured it out after a while.

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These are in fact challenging problems.