REAL TEST 4 |
SAMPLE FINAL EXAM FOR AU 12 |
QUIZ 8 (ANSWERS) |
NOTES |
SAMPLE TEST2 |
SAMPLE TEST3 |
READ ALL EXAMPLES. SOME PROBLEMS ARE JUST LIKE AN EXAMPLE. |
CH. 8 |
CH. 8 MULTIPLE CHOICE TBA - Exercise/problems: General Theory of Collisions and Momentum Conservation: 2, 8 , 14 (ICQ-do not turn in), 13, 18, Inelastic collisions 22 (ICQ), 63, 74, 29, elastic collisions 31, 34( ICQ--try 76)), 35, 76*, 18 (check if elastic) , Impulse = Force*time = change in momentum 40 (try 44 !) center of mass 46 (try 52 and 51) |
TURN IN: General Theory of Collisions and Momentum Conservation: 2 (b) and (c) only, 8 , 13, Inelastic collisions 63, 74, 29, elastic collisions 31, 35, 18 (check if elastic), Impulse = Force*time = change in momentum 40 (try 44 !) center of mass 46 (try 52 and 51) |
* DISCUSSIONS PROVIDED. |
SELECTED DISCUSSIONS TO exercises/problems
BELOW.
ALL PROBLEMS ARE DUE EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES. |
2.
(b) |
8.(a) sum of Pi_{x} = sum of P_{fx} . 0 = (725/g)*(1.25 m/s) (N) +(625/g)*V, where V is the x-component of momentum of lighter skater. Note V <0. (b) Compute (1/2)*(725/g)*(1.25 m/s)^{2} + (1/2)*(625/g)*V^{2} . |
13. sum of Pi_{x} = sum of P_{fx}
. (0.250 kg)*V + 0 = (0.250 kg)*( -0.120 m/s) + (0.350)*(0.650 m/s) , where the initial V is the x-component of momentum of lighter puck. Note V > 0. |
63. This is a 1D problem of conservation of
momentum occurring at bottom of swing on vine. The collision
occurs horizontally along an instantaneous x -axis. We equate the total
x directed momentum just before and just after r the
collision. M*V + 0 = MV_{f} + mV_{f} since stuntman and
mad villain have the same common velocity just after the collision; they
remain interlocked at a common decreasing velocity during the entire
time they slide. But just before and just after the collision we
have: M*V = (M + m)*V_{f} where V the initial x-component of momentum of stuntman is found using conservation of energy: Mgh = (1/2)*M*V^{2}. Find V > 0. (c) Use conservation of energy (1/2)*(M + m)*V_{f} 2 = Heat. Heat = f_{k}*D = (0.250)*(m+M)*g , where f_{k} = 0.250*N , where N = mg on the flat surface with no extermal vertical component of force. Find D. |
74. See example 8.7. m*V = (m + M)*V_{f},
where V the initial x-component of momentum of bullet before
impact. Now use conservation of energy (1/2)*(m + M)*V_{f}^{2 } = (m + M)*gh, where h = L*(1- cos 34.7). Solve these two equations simultaneously for V_{f} and V |
29. See example 8.8 . |
31. See example 8.9 . (0.300 kg)*(0.80 m/s) = (0.300 kg)*V + (0.150 kg)*V ' and 0.80 m/s = V' - V. Solve these two equations simultaneously for V and V'. |
35 (b) (2.00 kg)*(2.00 m/s) = (2.00 kg)*V + (10.0 kg)*V ' and 2.00 m/s = V' - V. Solve these two equations simultaneously for V and V' (a) At the moment when the spring is at maximum compression the speed of the cars are equal and is given by V_{f}. Solve: m*V = (m + M)*m*V = (M + m)*V_{f}, where V = 2.00 m/s and m = 2.00 kg and M = 10.00 kg. Find V_{f} . The use conservation of energy: (1/2)*m*V^{2} = (1/2)*(m + M)*V_{f}^{2} + U_{s}. where U_{s} is the potential energy of the effective spring representing the bumpers at maximum compression. |
18. (a) See example 8.11. Even though this problem is not necessarily involving an elastic collision, the same method of solution applies. (0.400 kg)*(0.250 m/s) = (0.400 kg)*(0.200 m/s)*cos 36.9 + (0.600 kg)*V*cos theta and 0 = 0.400 kg)*(0.200 m/s)*sin 36.9 - (0.600 kg)*V*sin theta. Solve these two equations simultaneously for V and theta. Isolate V cos theta and V sin theta and divide the two equations to solve for tan theta and theta. Then go back and find V. (b) The kinetic energy of the system before the collision is the kinetic energy of the 0.400 kg mass since the other object is initially at rest. Compute this initial kinetic energy. Then compute the final system (total) kinetic energy = (1/2)*(0.400 kg)*(0.200 m/s)^{2} + (0.600 kg)*V^{2}. Compare the two kinetic energies. |
40 Y component of impulse = Y component of change in momentum = m*V_{fy} - m*V_{iy}. The initial velocity of the ball is zero and the final y component of velocity is found by solving 0 = (V_{fy})^{2} - 2*g*(5.50 m). |
46 (a) Xcm = [(9.00 kg)*1.5 * (7.5
kg)*(2.4)]/(16.5 kg). (b) Repeat with the right end at x = 0. |