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A Car of Mass M is at Rest on a Frictionless Horizontal Surface and a Pendulum Bob of Mass M Hangs from the Roof of the Cart. the String Breaks, the Bob Falls on the Floor, - Physics

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प्रश्न

A car of mass M is at rest on a frictionless  horizontal surface and a pendulum bob of mass m hangs from the roof of the cart. The string breaks, the bob falls on the floor, makes serval collisions on the floor and finally lands up in a small slot made in the floor. The horizontal distance between the string and the slot is L. Find the displacement of the cart during this process.

योग

उत्तर

The mass of the bob is m.
The mass of the cart is M. 

Considering the bob falls at point A.

Initial distance of centre of mass of the system from P is given as
\[x = \frac{m \times L + M \times 0}{M + m} = \frac{m}{M + m}L\]



When the bob falls in the slot, the distance of centre of mass of the system from P becomes zero.

\[\therefore \text{ Shift in centre of mass } = 0 - \frac{mL}{M + m}\]

\[ = - \frac{mL}{M + m} \text{ towards left}\]

\[ = \frac{mL}{M + m} \text{ towards right}\]

Therefore, the cart moves a distance of
\[\frac{mL}{M + m}\] towards right.

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अध्याय 9: Centre of Mass, Linear Momentum, Collision - Exercise [पृष्ठ १६०]

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एचसी वर्मा Concepts of Physics Vol. 1 [English] Class 11 and 12
अध्याय 9 Centre of Mass, Linear Momentum, Collision
Exercise | Q 13 | पृष्ठ १६०

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