You are standing on a large sheet of frictionless ice and holding a large rock. In order to get off the ice, you throw the rock so it has velocity 12.0 m/s relative to the earth at an angle of 35.0o above the horizontal. If your mass is 70.0 kg and the rock’s mass is 3.00 kg, what is your speed after you throw the rock? (See Discussion Question Q8.7.) Q8.7 A woman holding a large rock stands on a frictionless, horizontal sheet of ice. She throws the rock with speed v0 at an angle ? above the horizontal. Consider the system consisting of the woman plus the rock. Is the momentum of the system conserved? Why or why not? Is any component of the momentum of the system conserved? Again, why or why not?
Solution 28E Step 1 : Data given Velocity of the rock v =r12.0 m/s Mass of rock m = 3.00 kg r 0 Angle along the horizontal = 35.0 Mass of woman m = 70wkg We need to find the velocity of the velocity after throwing the rock v w Using law of conservation of momentum we have m v = m v w w r r Let us find the velocity the velocity of the rock along x axis It is given by v = v × sin rx r Substituting values we get v rx12.0 m/s × sin 35.0 0 v = 6.88 m/s rx Similarly velocity along y axis is v ryv × ros Substituting values we get 0 v ry12.0 m/s × cos 35.0 v ry9.82 m/s Hence we get the velocity of the woman as m w w m v r r mr( rx+ry)/2 v w m w 3.0 kg( ( 6.88 m/s+9.82 m/s) )/2 vw= 70 kg vw= 25.05m/s kg /70 kg v = 0.35 m/s w Hence the velocity of the woman after throwing away the rock is found to be 0.35 m/s