Question

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute as shown in (Figure 1)

Answer 1

moh u = √58.8 mys Potential energy of mans m a kinatte kinetic energy energy of wan in h = my2 regn - V 20 98x3 - J58 8 whe my=m my aan Ugao. If they stick after collision, het reloaty after colleton be up From momentum conservatron, mu + my up a (m, + m2) up mx 58.8 + 2m xo - (meam up up - mx √58.8 - 2.556 03860 2 mals 3 м up = 8.556 mis Tv = 2.556 m/s 3) If collision a elaste mom my am = √58.8 m/s, Ug = 0 - = my m2 , + 2m2 met my = 2 mi mit ut my mai mitmy we required relocity of mass m, re. Vi V m -9m 1588 - 55- 8 5318 m+am to u,. -ve sign because directions in opposite Again from energy conservation, wigh = ? ne = v.% (1588) 2x9.8 -hi 0.333 ml h = 0.33m