Question

Question no. 7 on
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Answer 1

Answer on Question #56597, Physics Mechanics Relativity

5.14 A stick of length $l$ lies on horizontal table. It has a mass $M$ and is free to move in any way on the table. A ball of mass $m$ , moving perpendicularly to the stick at a distance $d$ from its center with speed $v$ collides elastically with it as shown in figure - 5.117. What quantities are conserved in the collision? What must be the mass of the ball so that it remains at rest immediately after collision.

Solution

In the collision, linear momentum of the system (stick+ball), angularmomentum and kinetic energy are conserved. Using the law of conservation of angular momentum

$$m v = M V$$

By the law of conservation of angular momentum

$$m v d = J \omega$$

where $J = \frac{1}{12} M l^2$

From the principle of conservation of kinetic energy

$$\frac {1}{2} m v ^ {2} = \frac {1}{2} J \omega^ {2} + \frac {1}{2} M V ^ {2}$$

Then

$$m v ^ {2} = J \cdot \frac {m ^ {2} v ^ {2} d ^ {2}}{J ^ {2}} + M \frac {m ^ {2} v ^ {2}}{M ^ {2}} \Rightarrow m = \frac {m ^ {2} d ^ {2}}{\frac {1}{12} M l ^ {2}} + \frac {m ^ {2}}{M} \Rightarrow$$ $$l = \frac {12 d ^ {2} m}{M l ^ {2}} + \frac {m}{M} \Rightarrow m = M \left(\frac {l ^ {2}}{l ^ {2} + 12 d ^ {2}}\right)$$

Answer: $m = M\left(\frac{l^2}{l^2 + 12d^2}\right)$

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