course Phy 231 ??????y????assignment #023
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13:54:36 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 --> OK
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13:54:41 ** 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 --> OK
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17:16:30 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 --> I have the 11th edition, when I do the corrected response, I will do the whole problem.
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17:31:42 ** 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 --> dsy=+8m, ay=-9.8m/s/s, v0y=v0*sin(60deg)=.5v0. dsx=18m, ax=0, v0x=v0c0s(60deg)=.87v0. x=.5v0*t y=.87*v0t-.5gt^2 t=x/.5v0. y=.87*v0*(x/.5v0) - .5*g*(x/.5v0)^2 v0^2*y=.87v0^2*(2x) - .5*g*4x^2 v0^2*y - 1.73*v0^2*x=-2g*x^2 v0^2*(y-1.73x)=.2gx^2 v0=sqrt(-2g*x^2/(y-1.73x)), y=8m, x=18m. v0=sqrt(277m^2/s^2)=16.7m/s. t=2x/v0, t=2*18m/16.7m/s = 2.2s. vx=dx/dt=.5v0=.5*16.7m/s=8.35m/s. vy=dy/dt=14.4m/s-9.8m/s/s*t =14.4m/s-9.8m/s/s*2.16s = -5.6m/s. v at impact: sqrt((8.25m/s)^2 + (-5.6m/s)^2)=10m/s. angle= arctan(vy/vx)=-3.4deg.
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