#$&* course Phy 231 3/13 11am 019. `query 19
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Given Solution: `a** STUDENT RESPONSE: x component of the vector = magnitude * cos of the angle y component of the vector = magnitude * sin of the angle To get the magnitude and angle from components: angle = arctan( y component / x component ); if the x component is less than 0 than we add 180 deg to the solution To get the magnitude we take the `sqrt of ( x component^2 + y component^2) ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qExplain what we mean when we say that the effect of a force is completely equivalent to the effect of two forces equal to its components. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To find the effect of a force, we consider the simultaneous effects of the two component forces along an x- and y-axis as if they were occurring separately. If given those components, we can calculate the total magnitude and angle of the net force (and vice versa). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** If one person pulls with the given force F in the given direction the effect is identical to what would happen if two people pulled, one in the x direction with force Fx and the other in the y direction with force Fy. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qExplain how we can calculate the magnitude and direction of the velocity of a projectile at a given point. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we know the initial horizontal and vertical velocities prior to becoming airborne, we have a clear picture of the object's motion due to the acceleration of gravity. We can calculate total velocity the same we can calculate total force, using x and y components and trigonometry. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Using initial conditions and the equations of motion we can determine the x and y velocities vx and vy at a given point, using the usual procedures for projectiles. The magnitude of the velocity is sqrt(vx^2 + vy^2) and the angle with the pos x axis is arctan(vy / vx), plus 180 deg if x is negative. ** STUDENT QUESTION this says that there are the magnitude and the angle with respect to the positvie x aixs, I am not quite clear on this ar ethey added together? INSTRUCTOR RESPONSE If an object is thrown straight up in the air, its initial velocity is all in the vertical direction. Its angle as measured from the horizontal x axis is 90 degrees. It has no horizontal velocity; the horizontal component of its velocity is zero. In this case our calculations would verify the obvious: cos(90 deg) = 0, so the x component of the velocity is v_x = v cos(90 deg) = v * 0 = 0. sin(90 deg) = 1, so the y component of the velocity is v_y = v sin(90 deg) = v * 1 = v. If an object is thrown in the horizontal direction, its angle with the horizontal is 0 degrees. Its velocity is wholly in the horizontal direction. The vertical component of its velocity is zero. Our calculations again verify this: cos(0 deg) = 0, so the x component of this velocity is v_x = v cos(0 deg) = v * 0 = 0. sin(0 deg) = 1, so the y component of this velocity is v_y = v sin(0 deg) = v * 1 = v. Now if an object is thrown at some nonzero angle with horizontal, as it typically the case, the magnitudes of its velocity components are less than the magnitude of its velocity. For example an object thrown at angle 45 degrees, halfway between the direction of the x axis and that of the y axis, has equal x and y components. Our calculation verifies this cos(45 deg) = .71, approx., so the x component of this velocity is v_x = v cos(45 deg) = v * .71 = .71 v. sin(45 deg) = .71, so the y component of this velocity is v_y = v sin(45 deg) = v * .71 = .71 v. An object thrown at 30 degrees, closer to the direction of the x axis that to that of the y axis, has a velocity component in the x direction which is greater than that in the y direction. Our calculation will verify this: cos(30 deg) = .87, approx., so the x component of this velocity is v_x = v cos(30 deg) = v * .87 = .87 v. sin(30 deg) = .50, so the y component of this velocity is v_y = v sin(30 deg) = v * .50 = .50 v. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This makes me think of an Etch-a-Sketch. To get an angled line, you have to turn the horizontal and vertical knobs simultaneously. To increase or decrease the steepness of the line (the angle with the x-axis), you have to turn the vertical knob either faster or slower in relation to the horizontal knob. The result, if you're good, is a smooth line that doesn't look like a stairstep series of vertical and horizontal lines. ------------------------------------------------ Self-critique rating:
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Given Solution: `a** Initial vel in the x direction is v cos(theta), where v and theta are the magnitude and the angle with respect to the positive x axis. Initial vel in the y direction is v sin(theta). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `qUniv. 8.63 (11th edition 8.58) (8.56 10th edition). 40 g, dropped from 2.00 m, rebounds to 1.60 m, .200 ms contact. Impulse? Ave. force? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This is 8.67 in the 13th edition. The final velocity just before impact is sqrt(v0^2+2a`ds), or sqrt(2*9.8m/s^2*2m), or 6.3m/s. That means the momentum at that instant is 6.3m/s * .040kg, or .25kg*m/s. For the velocity just after impact, we calculate: vf^2=v0^2+2a`ds 0=v0^2+(2*-9.8m/s^2*-1.6m) 0=v0^2+31.36m^2/s^2 -v0=sqrt(31.36m^2/s^2) -v0=+-5.6m/s since we defined the downward direction as positive, this post-impact velocity must be negative. Momentum=-5.6m/s*.040kg =-.224kg*m/s So the momentum changes from .25kg*m/s to -.224kg*m/s, for a total change in momentum, or impulse of -.474kg*m/s. Fave*`dt=`dp Fave * .2 ms * (1s/1000ms) = -.474kg*m/s Fave= -.474kg*m/s / .0002s Fave=-2370kg*m/s^2 Fave=-2370N confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** You have to find the momentum of the ball immediately before and immediately after the encounter with the floor. This allows you to find change in momentum. Using downward as positive direction throughout: Dropped from 2 m the ball will attain velocity of about 6.3 m/s by the time it hits the floor (v0=0, a = 9.8 m/s^2, `ds = 2 m, etc.). It rebounds with a velocity v0 such that `ds = -1.6 m, a = 9.8 m/s^2, vf = 0. This gives rebound velocity v0 = -5.6 m/s approx. Change in velocity is -5.6 m/s - 6.3 m/s = -11.9 m/s, approx. So change in momentum is about .04 kg * -11.9 m/s = -.48 kg m/s. In .2 millliseconds of contact we have F `dt = `dp or F = `dp / `dt = -.48 kg m/s / (.0002 s) = -2400 Newtons, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): In trying to train myself to look at energy levels rather than equations of motion, I attempted several times to approach this by starting with the gravitational PE and thinking about changes in PE and KE. Would there be any way to do it that way, or is this a case where we really should stick with equations of motion? 10:55 AM 3/13/2015Self-critique rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
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Given Solution: `a** You have to find the momentum of the ball immediately before and immediately after the encounter with the floor. This allows you to find change in momentum. Using downward as positive direction throughout: Dropped from 2 m the ball will attain velocity of about 6.3 m/s by the time it hits the floor (v0=0, a = 9.8 m/s^2, `ds = 2 m, etc.). It rebounds with a velocity v0 such that `ds = -1.6 m, a = 9.8 m/s^2, vf = 0. This gives rebound velocity v0 = -5.6 m/s approx. Change in velocity is -5.6 m/s - 6.3 m/s = -11.9 m/s, approx. So change in momentum is about .04 kg * -11.9 m/s = -.48 kg m/s. In .2 millliseconds of contact we have F `dt = `dp or F = `dp / `dt = -.48 kg m/s / (.0002 s) = -2400 Newtons, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): In trying to train myself to look at energy levels rather than equations of motion, I attempted several times to approach this by starting with the gravitational PE and thinking about changes in PE and KE. Would there be any way to do it that way, or is this a case where we really should stick with equations of motion? 10:55 AM 3/13/2015Self-critique rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
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Given Solution: `a** You have to find the momentum of the ball immediately before and immediately after the encounter with the floor. This allows you to find change in momentum. Using downward as positive direction throughout: Dropped from 2 m the ball will attain velocity of about 6.3 m/s by the time it hits the floor (v0=0, a = 9.8 m/s^2, `ds = 2 m, etc.). It rebounds with a velocity v0 such that `ds = -1.6 m, a = 9.8 m/s^2, vf = 0. This gives rebound velocity v0 = -5.6 m/s approx. Change in velocity is -5.6 m/s - 6.3 m/s = -11.9 m/s, approx. So change in momentum is about .04 kg * -11.9 m/s = -.48 kg m/s. In .2 millliseconds of contact we have F `dt = `dp or F = `dp / `dt = -.48 kg m/s / (.0002 s) = -2400 Newtons, approx. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): In trying to train myself to look at energy levels rather than equations of motion, I attempted several times to approach this by starting with the gravitational PE and thinking about changes in PE and KE. Would there be any way to do it that way, or is this a case where we really should stick with equations of motion? 10:55 AM 3/13/2015Self-critique rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: