Question

Part : Problem Solving (12 points each) Instructions: Solve all four problems. 1. The Schwerer Gustav was a huge railway gun used in wW II. The railway gun had an unloaded mass of 1.35 x 10° kg and fired shells with a mass of 7000 kg. If both the gun and the shell were stationary to start and the shell is fired at 820 m/s, what is the recoil velocity of the gun? 2. A 850 kg car initially going 18.0 m/s only in the x-direction runs into a stationary 1400 kg truck. After the collision the car is going 11.0 m/s at an angle of 38 degrees above the x-axis. What is the magnitude and direction of the velocity of the truck right after the collision (give speed and angle)? 3. A car initially traveling at 29.0 m/s undergoesa constant negative acceleration of magnitude 2.20 m/s. If the tires have a radius of 0.330 meters, how many revolutions does each tire make before the car comes to a stop? A bug hangs on by friction to a horizontal spinning disk at a radius of 28.2 cm. If the coefficient of friction between the bug and the disk is 0.880 (a) at what angular velocity must the disk spin to cause the bug to slide off? (b) Find the answer in degrees per econd. (c) Find the answer is revolutions per second.

Answer 1

TE Soln (1) m=1.35x106kg, m=7000kg u=82omls. from Conservation of momentum murmv v=mu= 7000 4820 = 4.252 ml 1.35106' V=4.252 mls.) m=850k9, U, - 18 mls, mon=1400 kg, Un=0, Villmi. Inihal momentum tim, uitma ug) î = 850x18=15300 kgols i onmentem je 1400 Final momentum, B = (miv, cos980+ more cosc) â + (mi visin38'- rov va Sinco) ſ droom Conseovahon of momentem miv, singer-M2 V2 since=0 V2 3166 = mi v, Sin38* = 850_XII Sin3 8° V ₂ Sino=holl (1) m, v, Cos38+ ma vo Cos0 – 15300 Do 1400 v2 cover 7932.09 huolto V2CO3 0 = 5.67—(1) Squaring and adding ein (1) and (1), we get 2 = 49.ou1 [V2 = 7 m/s a speed of truck :: Sinc = 410l1 = 0.587 7 [0=35.95° below horizontal

) U= 29 mle, a= 2.20 m 112,8 = 0.33m , WAV Wo= u. 87.88 vaale, d = Q = 6.67 radle2 w2_o2=-220 0 = W.? =( 8812 L. 578.93 rad. 2x6.67 * In = 92.14 a vevons. 2 o 2.14 revoin. (4) R = 0.282 m, MK = 0.880 de moar mw2R = Mimg W= Vuka - J 0.880x9.8 -5.53 ad s. U=5.53 radli. VR 0.282 w = 5.53x180° - 316.85 B E6.00 18.] W=20v -= 0.880 gps. V = W