Question

(6%) Problem 6: The figure shows the view looking down onto an air hockey table. A puck collides with a stationary bar. The collision is completely inelastic. 1 The bar has mass M= 1.88 kilograms and length L = 1.47 meters. The puck has mass m = 0.466 kilograms and is initially moving at v; = 6.83 perpendicular to the bar. The puck collides with the bar at a location of b = 0.475 meters from the center of the bar. Note that the size of the puck is negligible compared to the length of the bar. - - - M . ►A 50% Part (a) Calculate the center of mass speed (in m/s) of the puck-and-bar system just after the collision. Vi= 1 Grade Summary Deductions 0% Potential 100% HOME E sin() cos cotan) asin atan acotan cosh tanh Degrees tant acos sinh() cotanh0 Radians ( ) 7 8 ^^^ 4 5 * 1 2 + - 0 VO BACKSPACE 9 6 3 . DEL Submissions Attempts remaining: 15 (2% per attempt) detailed view → END CLEAR Submit Hint Feedback I give up! Hints: 0 for a 0% deduction. Hints remaining: 0 Feedback: 2% deduction per feedback. A 50% Part (b) Calculate the angular speed (rad/s) about the center of mass of the puck-and-bar system just after the collision.

Answer 1

M = 1.88 kg, m = 0.466kg. L = 1.47m. V -6.83 m/s. b=0.475m. and I Veow (Initial velocity diogram) (Impube diagrous) Net Impure diagram (zero Impuhe) - Here het Tupute on system is zero. Hence Momenten of erystuen is Conserved mvi - me, + MVcom o w speed of buck After Collision Vcom - speed of Centre of mass of ber After Collision [fy-y] The collision is perfectly inelastic Hence Verepe Vapp - Veep = 0. Vab. Velocity of velocity of seperation) Approch) and V is the velocity of beer at the live of impact ( Red Colour)

or Vop Hence & P - Vi = 0.vi angular moment un of the system is also Consene about Centre of mass of the bar bcoz net Impure is zero. by Aguler Momenten conservation mi Li= Lp mvib- ml.bt Iew. wt ② Tana Mh² for bar 12 How assume w is the Anguler velocity of cooM if rod. and solving all @ghs. from en 0.466 X 6.83 = 0.466 4 + 1.88 Wom [ 0.4662, +1.88 Vacan, 3.18 6 from en ③ 0.466x68340.478 0.466 x V x 0.47 + 188x(47) 12

10.22 V, +0.712 4 =11511) TV2 = Veomt wb 1 comt w b=20, from egn ③ 15815 0.2220, it 1.88x6:47)2 w. 1.51l = 0.22 , + 0.338W W = 115 11- 0.2224, 0. 338 8 Ver + (1.511-0.228, 10-415 = 8, 0.338 = 2, Vean + 1.4 (151-0-220) = 20, Nem + 2.122-0-3084, = 0, Iven = 1:3084,-2.1227 Now but it in egr 2 0.466, +1.88 (1 3080,- 2.122) = 3.18 oubore, +2.4610, – 3.88 = 3.18 2.9262 = 7:16 Fre, = 2.40 m/s)

Hence Van= 930828,- 2.122 Van = 1:078 m/s ų = 28, = 2.44 m/ fur enn w Von V W = V.- Veon 2.441-078 - = 2.86 rady Part @ Centre of mass speed - Veom a mol + M Vean 0466x2 44+ 1.888 1078 +O4CC Veom ww = 1.348 ms. An Part 6) - we have already Calculated about C.O.M W = 2.86 rad's Ang conguher speed