Question

(5) Gayle runs at a speed of 4.49 m/s and dives on a sled, which is initially at rest on the top of a frictionless snow.covered hill. After she has descended a vertical distance of 5.46 m, her brother, who is initially at rest, hops on her back and together they continue down the hill. What is their speed at the bottom of the hill if the total vertical drop is 16.6 m2 Gayle's mass is 51.4 kg, the sled has a mass of 5.40 kg and her brother has a mass of 31.4 kg.

Answer 1

m, = mass of Gayle = 51.4 kg m. mass of sled z 5.40 kq mg = mass of her brother = 31.4 kg v = speed of Gayle = 4.49 m/s when Gayle dives onto sled, by conservation of momentum, we have (m, & m2) V₂ = mivi t mrvo - ₂ = speed Vo = speed of (Gayle + sled) together of sled = o . Va =_ me V mitmz 51.4 *441 51.4 +5.4 :.V = 4.06 m/s when Gayle & sled have undergone vertical distance of an 5.46 m, by conservation of chersy, we have 1 (mm) 42 + 0 (mm) g (6 h) - mythes von 3. Vo = √ ² + eg (hz - hi) 6.06)2+ 2x9.8 x 5.46m - V3 = 11.11 m/

When her brother, initially at rest , hops on her back, conservation of momentum gives (m, tm tms) v = (m, + m2) V₂ + m₂ (0) m, tmz mitm₂+ m₂ 51.4+5.40 51.4+5.40+ 31.4 She said * Hett V = 7.15 m/s finally, after combined system undesgone an additional (16.6-5.46) = 11.14 m drop then be consenation of mech. enessy gives - simptomat my bus įrm that man kan - + (,1 met mang the host :: VG = V Vp + 29 (he thr) = V 7.152 + 2x9.8 X 11.14 m TVs = 16.4 ml - Answer