- #1

- 36

- 0

- Homework Statement
- Two material points of identical mass are in free fall, connected by a rigid rod of negligible mass and length ##L##. At time ##t=0## both masses have zero initial velocity and are positioned at different heights above the ground, the rod forming an angle ##\theta## with the horizontal. Determine how the system moves immediately after the first contact with the ground, imagining that the ground is perfectly smooth and assuming a fully elastic collision model for the impact.

- Relevant Equations
- /

The system of two material points of identical mass connected by a rigid rod of negligible mass and length ##L## is an example of a conservation of energy problem. The initial energy of the system is the sum of the kinetic energy of the two points and the potential energy of the rod, which is zero since it is horizontal. The final energy of the system is also the sum of the kinetic energy of the two points and the potential energy of the rod, which is zero since it is vertical. Therefore, we can write: ##E_i=E_f## $$1/2mv_i^2+mgh_i+mgh_f =1/2mv_f^2+mgh_f$$ where ##m## is the mass of each point, ##v_i## and ##v_f## are their final velocities, ##g## is the gravitational acceleration, ##h_i## and ##h_f## are their initial and final heights above the ground, and ##h_f=L\theta##, where ##\theta## is the angle between the rod and the horizontal. Since we assume a fully elastic collision model for the impact, we can use conservation of momentum to relate ##v_i## and ##v_f##. The initial momentum of the system is: $$p_i=mv_i+mgh_i=mL\cos\theta + mgh_i$$ The final momentum of the system is: $$p_f=mv_f+mgh_f=mL\cos\theta+mL\sin\thetag+mgh_f =mL(\cos\theta+\sin\theta)g+mgh_f =mL(\cos\theta+\sin\theta)g−mL(\cos\theta+\sin\theta)g=mL(\cos 2\theta-\sin 2\theta)g=mL(0)g=0$$ Since momentum is conserved, we have: $$p_i=p_f \Rightarrow \ mvi +mghi=0 \Rightarrow \ v_i=−mg/L \Rightarrow v_ f=mg/L \Rightarrow v_f=v_i$$ Therefore, we can substitute this into our energy equation and get: $$E_i=E_f \Rightarrow 1/2mv_i^2 +mgh_i+mgh_f=1/2mv_f^2+mgh_f \Rightarrow h_f-h_i−L \theta g=0 \Rightarrow h_f - h_i - L \theta g=0$$ This means that after the first contact with the ground, both points will have zero height above it. The rod will remain vertical at an angle ##L \theta##. Where do I go wrong? Could you help me understand the physical situation? Thank you.