Question

Show major steps and formulas involved!!!!!, really need to understand how to get the results, if you can draw something about tge picture, I will appretiated!!!!!

80. A small block of mass m = 0.500 kg is released from W rest at the top of a frictionless, curve-shaped wedge of mass m, = 3.00 kg, which sits on a frictionless, hori- zontal surface as shown in Figure P9.80a. When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right as shown in Figure P9.80b. (a) What is the velocity of the wedge after the block reaches the horizontal surface? (b) What is the height h of the wedge? any 1712 Figure P9.80

Answer 1

m = o.sookg m2 = 3.00kg Vi= umis so momentum is zero @ Initially both are ad rest mv, + M2 V₂ final momentum pro Pf = 0.54 4 + 3 V2 from law momentum conservation P = PF of 2 + 3V V2 = – 2 = -0.67 m/s in left direction Now wing energy conservation migh = I minh + 1 M2 V2² 0.5x9. 8 th= (x 0.54 (4)2 + 1 + 3 (067) ² yogh = 4 +0.673 yigh = 4.673 o.gsm h = 4.673 = 4.g I h = 0.95m 1