#$&* course Phy 231 022. `query 22
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Given Solution: The bears exert equal and opposite forces on one another, which act for equal time intervals. If no forces act other than these contact forces, this results in equal and opposite changes in momentum. The collision takes place on ice, so frictional forces will be small and for the short interval of the collision may be disregarded. The total momentum of the two-bear system therefore remains constant during the collision. The total momentum after collision is therefore equal to the total momentum before. Before collision the second bear is stationary, so the total momentum is just the (.2 kg * .75 m/s) = .15 kg m/s momentum of the first bear. After collision the two bears constitute a mass of .2 kg + .35 kg = .55 kg, and we do not yet know their common velocity. If we let u stand for their common unknown velocity, their momentum after collision is therefore momentum after collision = .55 kg * u. Since the momentum is the same after collision as it was before, we therefore have .15 kg m/s = .55 kg * u so that u = .15 kg m/s / (.55 kg) = .7 m/s, approximately. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I do not understand how you got .7. .15/.55 is .27272??? ------------------------------------------------ Self-critique rating:
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Given Solution: `a** We'll take East to be the positive direction. The original magnitude and direction of the momentum of the fullback is p = m * v1 = 115kg (4m/s) = 380 kg m/s. Since velocity is in the positive x direction the momentum is in the positive x direction, i.e., East. The magnitude and direction of the impulse exerted on the fullback will therefore be impulse = change in momentum or impulse = pFinal - pInitial = 0 kg m/s - 380 kg m/s = -380 kg m/s. Impulse is negative so the direction is in the negative x direction, i.e., West. Impulse = Fave * `dt so Fave = impulse / `dt. Thus the average force exerted on the fullback is Fave = 'dp / 'dt = -380 kg m/s /(.75s) = -506 N The direction is in the negative x direction, i.e., West. The force exerted on the tackler is equal and opposite to the force exerted on the fullback. The force on the tackler is therefore + 506 N. The positive force is consistent with the fact that the tackler's momentum change in positive (starts with negative, i.e., Westward, momentum and ends up with momentum 0). The impulse on the tackler is to the East. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The horizontal velocity is unchanging so the horizontal component is always equal to the known initial horizontal velocity. The vertical velocity starts at 0, with acceleration thru a known distance at 9.8 m/s^2 downward. The final vertical velocity is easily found using the fourth equation of motion. We therefore know the x (horizontal) and y (vertical) components of the velocity. Using the Pythagorean Theorem and arctan (vy / vx) we find the speed and direction of the motion. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I forgot to remember that acceleration is a downward force due to gravity, so it should be -9.8 m/s^2. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qGive at least three examples of vector quantities for which we might wish to find the components from magnitude and direction. Explain the meaning of the magnitude and the direction of each, and explain the meaning of the vector components. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1. When are force is moving an object at an angle. 2. The way an object rotates around an a point or axis 3. The magnitude and direction of each of these examples is that velocity and direction of the object. The vector components are the x comp (horizontal) and y comp (verticle) which give the magniutude and direction of an object. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 2.`a Examples might include: A force acting on an object causing it to move in an angular direction. A ball falling to the ground with a certain velocity and angle. A two car collision; velocity and momentum are both vector quantities and both important for analyzing the collision.. The magnitude and directiohn of the relsultant is the velocity and direction of travel. The vector components are the horizontal and vertical components that would produce the same effect as the resultant. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): If did not mention the resultant and what it was. ------------------------------------------------ Self-critique Rating:2 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!