As shown below, a ball of mass
m, suspended on the end of a wire, is released from height
h and collides elastically, when it is at its lowest point, which a block of mass
2m at rest on a frictionless surface. After the collision, the ball rises to a final height equal to:
- \dfrac{h}{9}
- \dfrac{h}{8}
- \dfrac{h}{3}
- \dfrac{h}{2}
- \dfrac{2h}{3}
2 comments:
Post a Comment
This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.
One method I learned than solving the slightly complicated system of equations is to consider the center of mass frame (CM). If mass M falls down with velocity v in the CM frame, then it must return with velocity -v. Thus v_{CM}=\frac{Mv_1+2Mv_2}{M+2M} where v_1=\sqrt{2gh} by energy conservation. We thus see the return velocity,v_1, is -\frac{v}{3}. Using energy conservation again, the return height h_{new}=\frac{h}{9}
Yup. That's exactly the method I was referring to in the solutions. Good job, Matt. =]
Sorry for the bad LaTeX formatting on my part.
One method I learned than solving the slightly complicated system of equations is to consider the center of mass frame (CM). If mass M falls down with velocity v in the CM frame, then it must return with velocity -v. Thus v_{CM}=\frac{Mv_1+2Mv_2}{M+2M} where v_1=\sqrt{2gh} by energy conservation. We thus see the return velocity,v_1, is -\frac{v}{3}. Using energy conservation again, the return height h_{new}=\frac{h}{9}
Yup. That's exactly the method I was referring to in the solutions. Good job, Matt. =]
Post a Comment