QUIZ 10 - CH 8 - MOMENTUM AND ENERGY ( (SAMPLE TEST 3, FOR ENERGY) AND ( SAMPLE TEST 4 SPRING '11, FOR MOMENTUM ) (ANSWERS) |
8, 13, 22, 24, 28 (TRY 95), 31, 36, 41, 43 (TRY 82 )
, 46, 48, 49, 51, 54, 55 (TRY 51 AND 54 ), 57( TRY 107), 82 (TRY 43), 86. |
DISCUSSIONS POSTED BELOW: |
8. (a) IMPULSE = CHANGE IN MOMENTUM. See Example 8.2
and also note this problem is related to the last lab-- A VERNIER
LAB WILL HAVE A MINI LAB ON THIS SUBJECT. Note: m*v_{f} - m*v_{i} = change in momentum where v_{i} and v_{f} represent velocity components along the x - axis. Let rightward be positive. For example the pitched ball initially moves to the right, so v_{i} is positive and equals +45.0 m/s. After impact with the bat, the ball moves left in the negative direction; thus v_{f} < 0 and equals - 55.0 m/s . (b) Clearly the x-component of force F_{x} is negative. We have m*v_{f} - m*v_{i} = F_{x} *change in time. Note F_{x} is the average force. Normally, Impulse = m*v_{f} - m*v_{i} = area under the force curve, obtained via integration. But it can be shown that the area under the curve = average force *(change in time) from the MEAN VALUE THEOREM FOR INTEGRALS (Review Math 1) . |
13. (a) IMPULSE = CHANGE IN MOMENTUM. The problem seems to have been plucked from EXP 20 VERNIER where you had to analyze the shape of the force curve. In the experiment, clearly the shape was not a perfect rectangle like this idealization , but the main concepts apply: The area under the curve represents the change in momentum resulting from the force. | m*v_{f} - m*v_{i} | = area under the force curve. Note we have given you the magnitude of the change in momentum or impulse since the force curve plots the force magnitude. (b) Assume the rightward direction is positive. You are given v_{i} = +5.00 m/s. To get the exact signed component of impulse you would write: m*v_{f} - m*v_{i} = area under the force curve. (i) If the force acts rightward m*v_{f} - m*v_{i} = area under the force curve = positive area of rectangle, in which case the final velocity component will be greater than the initial velocity component. (ii) If the force acts leftward m*v_{f} - m*v_{i} = area under the force curve = negative area of rectangle, in which case the final velocity component will be less than the initial velocity component ; indeed v_{f } could be negative depending on the magnitude of the negative area but that is up to you to investigate |
22. m_{1}*v_{1i} + m_{2}*v_{2i}
= m_{1}*v_{1f} + m_{2}*v_{2f}
. Let right be positive; thus v_{1i} = + 1.50
m/s and v_{2i} = -1.10 m/s. Also you are given that
v_{1f} = + 0.250 m/s. (a) Find velocity component of the second (2) mass. Hint After collision the second mass is moving to the right. (b) There are two ways to explore this. One way is to find out if the collision is elastic or not and possibly bypass the kinetic energy computation. See equation 8.27 and discussions thereof and simply find out if the magnitudes of the relative velocities are equal before and after the collision. In other words , check if : v_{1i} - v_{2i} = v_{2f} - v_{1f} . Note we used this concept in EXP 19 Vernier for the magnetic bumper to magnetic bumper collisions. If the equality is true, then the collision is elastic and the change in the combined kinetic energy will be zero. If the equality is false, then you must compute the change in combined kinetic energy in the following way: Change in combined kinetic energy = (1/2)*m_{1}*v_{1f} ^{2 }+ (1/2)*m_{2}*v_{2f} ^{2} - (1/2)* m_{1}*v_{1i} ^{2} - (1/2)*m_{2}*v_{2i} ^{2} . |
24. (a) This part is essentially a replica of example 8.4. m_{A}*v_{Ai}
+ m_{B}*v_{Bi} = m_{A}*v_{Af}
+ m_{B}*v_{Bf} , where m_{A}*v_{Ai} + m_{B}*v_{Bi} = 0, since the closed system starts from rest. Thus, m_{A}*v_{Af} + m_{B}*v_{Bf} = 0 and you can now find the final velocity of A given B's final velocity. (b) Spring potential energy = (1/2)*k*x'^{2} = combined kinetic energy after the the spring has expanded = (1/2)*m_{A}*v_{Af} ^{2 }+ (1/2)*m_{B}*v_{Bf} ^{2 } . From this you can find the spring potential energy. Note if I gave you the spring's force constant k, you could find the distance |x'| it was compressed. |
28. This is a subtle variation of #24--- Here the system (
YOU AND THE ROCK) is also at rest initially but the rock has a
final velocity that is not collinear with the final velocity of
You as you slide in a direction opposite the horizontal component
of rock's final velocity. m_{1}*v_{1i} + m_{2}*v_{2i} = m_{1}*v_{1f} + m_{2}*v_{2f} . which translates into: 0 = m_{1}*v_{1f} + m_{2}*v_{2f} , where the written velocities are are x-components. Let 1 = You and 2 = rock, Clearly v_{2f} = 12*cos 35. Find v_{1f} . We are assuming rightward is positive, so you will recoil to the left in the negative direction. Note we have assumed for this system, rock and You, the net external force in the x direction is zero, resulting in conservation of momentum in the x-direction. In other words, in the x direction, the only relevatn forces acting on system members You and rock are action reaction pairs , which sum to zero. In the y direction , momentum not conserved for the system of You and rock since the rock has an upward y directed velocity of 12*sin 35 and You have NO y-component of velocity. This will be discussed in greater detail later, but the bottom line is this: The rock transfers its y-component of momentum to the Earth. So momentum is conserved in the y direction if you expand the system to include the Earth. In other words, it can be easily be shown m_{2}*12*sin 35 = momentum transferred to Earth. Here is how: ISOLATE THE ISOLATE YOU ( AND YOUR HAND) UNDER THE We have the following data to compare with regard to the
normal force of magnitude N' on Earth exerted by you: THE proof could repeated in the case of "tryout" # 87, discussed below. But in future problems a proof will normally not be required. |
31. See example 8.12. In the x direction momentum is
conserved: m_{A}*(40.0 m/s) = m_{A}*V_{A} cos 30 + m_{B}*V_{B}cos 45 In the y direction momentum is conserved: 0 = m_{A}*V_{A} sin 30 - mB*V_{B}sin 45. Solve these two equations simultaneously for V_{A} and V_{B}. Then answer the other questions. |
36. CLASS NOTES. |
41. See example 8.9. After they collide, the momentum is (m + M)*V_{f}* at an angle of 24 degrees with vertical or 66 degrees with the horizontal . We have (m + M)*V_{f} cos 66 = m*V_{c} and (m + M)*V_{f} sin 66 = M*V_{T} where m is the mass of the compact, M is the mass of the Truck and V_{f} = 16.0 m/s. Find Vc andV_{T}. |
43. CLASS NOTES. |
46. m_{1}*v_{1i} + m_{2}*v_{2i} = m_{1}*v_{1f} + m_{2}*v_{2f} and v_{1i }- v_{2i} = v_{2f} - v_{1f} . Solve the two equations simultaneously for the two final velocities. Let m_{1} = 0.150 kg and m_{2 }= 0.300 kg, where v_{1i} = +0.80 m/s and v_{2i } = -2.20 m/s. |
48. m_{1}*v_{1i} + m_{2}*v_{2i} = m_{1}*v_{1f} + m_{2}*v_{2f} and v_{1i }- v_{2i} = v_{2f} - v_{1f} . Solve the two equations simultaneously for the two final velocities. Then answer the other questions. Let m_{1} = 30.0 g and m_{2 }= 10.0 g, where v_{1i} = +0.200 m/s and v_{2i } = -0.400 m/s. |
49. m_{1}*v_{1i} + m_{2}*v_{2i} = m_{1}*v_{1f} + m_{2}*v_{2f} and v_{1i }- v_{2i} = v_{2f} - v_{1f} . Solve the two equations simultaneously for the two final velocities. Let m_{1} = 1.0 u and m_{2 }= 2.0 u, where v_{1i} = v and v_{2i } = 0. Your answers for the 2 velocities will be in terms of v. Then answer the other questions. |
51. CLASS NOTES. |
54. CLASS NOTES. |
55. CLASS NOTES |
57. See example 8.14. Since the closed system starts from rest, 0 = m_{1}*v_{1i} + m_{2}*v_{2i} = m_{1}*v_{1f} + m_{2}*v_{2f} . Thus, m_{1}*v_{1f} + m_{2}*v_{2f} = 0. What is v_{1f} if v_{2f} = -0.70 m/s ? |