Download Constant Acceleration - General Physics - Solved Past Paper and more Exams Physics in PDF only on Docsity! AJM:6/12/09 Page 1 of 1 Midterm Solutions Physics 131 Final Solutions Spring 2009 1. ! F 3 = m ! a ! ! F 1 ! ! F 2 = 15 N ! cos 30°x̂ ! sin 30°ŷ( ) ! 24 N cos 60°x̂ + sin 60°ŷ( ) ! 40 N sin 20°x̂ ! cos 20°ŷ( ) = !38.7 N!x̂ + 9.3 N!ŷ = 39.8 N, 166.5° this direction is 43.5° CW from the acceleration vector. 2. Solve the simple one-dimensional, constant acceleration problem to find that the acceleration is a = 2h t 2 = 2.5 m/s 2 . Then apply Newton’s second law to both blocks to find two equations involving the acceleration and the tension. Eliminate the tension and solve for m table to find that mtable = mhanging g / a !1( ) = 6.0 kg . 3. a) Apply conservation of momentum to find V x = 80 g( ) 20 m/s( ) ! 60 g( ) 20 m/s( ) 80 g = 5.0 m/s Vy = ! 60 g( ) 20 m/s( ) 80 g = !15.0 m/s Thus ! V = 15.8 m/s,!-71.6° . The direction is 71.6° CW from the final velocity of the 60 g rock. b) Clearly this was not an elastic collision because the 60 g rock has the same kinetic energy afterwards and the 80 g rock has less. EXTRA CREDIT By constructing the vector diagram for ! ! p for the 60 g rock, it’s easy to see that ! ! p is just 2 times the magnitude of its momentum before or after the collision. Thus, F = ! ! p !t = 2 60 g( ) 20 m/s( ) 8.0 ms = 212 N 4. Looking at the vertical motion during the first three seconds, show that vyi = g!t = 30 m/s . Thus, show that vxi = vi 2 ! vyi 2 = 40 m/s . Thus, looking at the horizontal motion during the entire flight, show that the total time of flight is !t = !x v xi = 8.0 s . Finally, looking at the vertical motion show that !y = "80 m . Thus, the cliff is 80 m high. 5. Apply Newton’s Second law to the car at the top of the loop to find that mv 2 r = .30 N + mg = .80 N and that, therefore, the kinetic energy is 1 2 mv 2 = 0.12 J . Then apply conservation of energy to the period from just before launch to the top of the loop to find that 1 2 kx 2 = 1 2 mv 2 + mg 2R( ) = .42 J and that, therefore, k = 2 .72 J( ) .02 m( ) 2 = 2100 N/m . EXTRA CREDIT Eliminating the 0.30 N contact force from step one above, we find mv 2 r = .50 N , 1 2 mv 2 = 0.075 J , 1 2 kx 2 = 1 2 mv 2 + mg 2R( ) = .375 J , so that x = 2 .375 J( ) 2100 N/m = 1.89 cm 6. a) Clearly, the two tensions have to add up to the weight of the beam, 1000 lb. b) Recognizing that the gravitational force acts at the center of mass, which will be at the center of the beam and taking torques around the right end of the beam we find ! net = +Fg L / 2( ) " Tleft 45 L( ) = 0 so Tleft = 5 8 1000 lb( ) = 625 lb and Tright = 1000 lb ! 625 lb = 375 lb
