course Phy231 5/11/10 3:30pm 19. Open Query019. `query 19
.............................................
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):ok ------------------------------------------------ Self-critique rating #$&*ok ********************************************* Question: `qExplain what we mean when we say that the effect of a force is completely equivalent to the effect of two forces equal to its components. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Your explanation says it all. I couldn't have said it any better. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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):ok ------------------------------------------------ Self-critique rating #$&*ok ********************************************* Question: `qExplain how we can calculate the magnitude and direction of the velocity of a projectile at a given point. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: given initial properties velocities for x and y can be determined magn='sqrt(vx^2+vy^2) angle=arctan(vy/vx), add 180deg is x is less than 0 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 are they 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):ok ------------------------------------------------ Self-critique rating #$&*ok ********************************************* Question: `qExplain how we can calculate the initial velocities of a projectile in the horizontal and vertical directions given the magnitude and direction of the initial velocity. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: initial velocity x=v cos(theta) v=magn theta=angle initial velocity y=v sin(theta) confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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):ok ------------------------------------------------ Self-critique rating #$&*ok ********************************************* Question: `qUniv. 8.63 (11th edition 8.58) (8.56 10th edition). 40 g, dropped from 2.00 m, rebounds to 1.60 m, .200 ms contact. Impulse? Ave. force? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v0=0m/s a=9.8m/s 'ds=2m vf=+-'sqrt(v0^2+2a'ds)=+-'sqrt(0+2*9.8m/s^2*2m)=6.3m/s vf=0m/s 'ds=-1.6m a=9.8m/s^2 vf^2=v0^2+2a'ds vf^2-2a'ds=v0^2 v0=+-'sqrt(vf^2-2a'ds)='sqrt(2*9.8m/s^2*-1.6m)=-5.6m/s 'dv=-5.6m/s-6.3m/s=-11.9m/s 'dm=.04kg*-11.9m/s=-.48kg m/s F'dt='dp F='dp/'dt=.48kg m/s/ (.0002s)=-2400N confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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):ok ------------------------------------------------ Self-critique rating #$&*ok"