Question

Constants In a shipping company distribution center, an open cart of mass 45.0 kg is rolling to the left at a speed of 5.50 m/s (see the figure (Figure 1)). You can ignore friction between the cart and the floor. A 19.0 kg package slides down a chute that is inclined at 37 from the horizontal and leaves the end of the chute with a speed of 2.80 m/s The package lands in the cart and they roll off together Figure 1 of 1 37° 4.00 m 25

Answer 1

ucoso Uomo giren u= 2.8 m/s h =4m. - - initial relocity, - LY u= ulos 37" ilusnost a = 2.8[cos37 - Sis 37) 22,815? - ) u = 2,844 ? - 33) = 2:24 1 - 1.689 let it be the velouty with which it hits the cart 5)= velo vg x = ucoso = 2,80837 = 2.814 ) = Penis ry= suyragh = f (usun37") + (2 X9,8x4) -|(2,5 43 ) + (2x9.8x4) 2,24 m/s N y = 9.012 m/s .:22= 2.24 Q_9.012 e velocuity with which it lands is V2 = (W1 = 12,242 -9,012 il Ilvel = $12, 24) + (9,012) = 9,286 m/s = 2 = 9,29 m/s

part B i let v be the common relocity attained. According to conservation of momentum, m. V + mva (m+m2) v = = m,v,' + my ve m, tm2 given m, = 45kg = 5.51-8) m₂ = 1969, = 2,248 _ 9,012 9. - 451-5,58) + 19{2,24?_9,012 3] 45+19 V =.-204,94 Â -171.228 § 64 - 151 = 4.1727 m/s ie Voot = 4.17 m/s ✓ Cart final