 #1
PiEpsilon
 22
 2
 Homework Statement:

Two uniform, equal stiff rods AB and AC are freely hinged at A and placed on a smooth horizontal table with AC perpendicular to AB, as shown in Fig. 162.
A horizontal blow is delivered perpendicular to AC at C. Find the ratio of the resulting linear velocities of the centers of mass of the rods Vac/Vab, immediately after this impulse.
 Relevant Equations:
 None
We know that impulse is
$$\vec J = \vec F \Delta t = \Delta \vec p$$
Let ##l, m## be the length of single rod and its mass respectively.
Analyzing torques and forces on each rod separately we have:
Rod ##AC##:
$$F\Delta t +N_x\Delta t = mV_{ac,x} \space\space\text{ eq. }(1)$$
$$F\Delta t\cdot \frac l2  N_x\Delta t\cdot \frac l2 = \frac{ml^2 \omega_{ac}}{12}\implies F\Delta t  N_x\Delta t = \frac{ml\omega_{ac}}{6}\space\space\text{ eq. }(2)$$
Rod ##AB##:
$$ N_x \Delta t = mV_{ab,x} \space\space\text{ eq. } (3)$$
$$\text{where } N_x \text{ is the force acting from the hinge connection}$$
Adding ##(1)## to ##1*(2)##, we have the following relation:
$$2N_x \Delta t = mV_{ac,x}  \frac{ml\omega_{ac}}{6} \implies 2V_{ab,x} + V_{ac,x} = \frac{ml\omega_{ac}}{6} \space\space\text{ eq. }(4)$$
Analyzing the situation from the CM view left me with no meaningful result.
Work energy consideration seemed also to lead nowhere.
Next logical step would be to represent right side of equation ##(4)## as some other relation of velocities of ## V_{ab,x} ## and ##V_{ac,x}## which I am unable to do.
I have left vertical direction, as vertical velocities of rods must be equal in magnitude but opposite in direction.*
*when the system of rods is considered, the force acting is horizontal.
Since ##\vec F = 2m\vec A_{cm} \implies \vec V_{cm}## is horizontal and:
##2m\vec V_{cm} = m\vec V_{ac} + m\vec V_{ab} \implies V_{cm,y} = 0 \iff V_{ac,y} = V_{ab,y}##
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The horizontal positive ##x## direction is taken as pointing to the right (from B to A)
The vertical positive ##y## direction is taken as pointing up (from C to A)
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