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Byju's Answer

Standard XII

Physics

COM in Collision and Explosion

In the figure...

Question

In the figure shown the cart of mass '6m' is initially at rest. A particle of mass 'm' is attached to the end of the light rod which can rotate freely about A. If the rod is released from rest in a horizontal position shown, determine the velocity $v_{r e l}$ of the particle with respect to the cart when the rod is vertical.

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Solution

The correct option is C

When the rod becomes vertical

$m v_{1} = 6 m v_{2} \Rightarrow v_{1} = 6 v_{2}$ .........(i)

$\frac{1}{2} m v_{1}^{2} + \frac{1}{2} 6 m v_{2}^{2} = m g l$ .........(ii)

Solve (i) and (ii) to get $V_{r e l} = v_{1} + v_{2} = \sqrt{\frac{7}{3} g l}$

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Similar questions

In the figure shown the cart of mass 6m is initially at rest. A particle of mass m is attached to the end of the light rod which can rotate freely about A. If the rod is released from rest in a horizontal position shown, determine the velocity $v_{r e l}$ of the particle with respect to the cart when the rod is vertical.

Q. In the figure shown, the cart of mass

6 m

is initially at rest. A particle of mass

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is attached to the end of the light rod which can rotate freely about A. If the rod is released from rest in a horizontal position shown, determine the velocity

v_{r e l}

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In the figure shown the cart of mass '6m' is initially at rest. A particle of mass 'm' is attached to the end of the light rod which can rotate freely about A. If the rod is released from rest in a horizontal position shown, determine the velocity $v_{r e l}$ of the particle with respect to the cart when the rod is vertical.

Q. A particle of mass

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and charge

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is attached to a light rod of length

L

. The rod can rotate freely in the plane of paper about the other end, which is hinged at

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. The entire assembly lies in a uniform electric field

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also acting in the plane of paper as shown. The rod is released from rest when it makes an angle

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Two beads of mass 2m and m, connected by a rod of length l of negligible mass are free to move in a smooth vertical circular wire frame of radius l as shown. Initially the system is held in horizontal position (as shown in the figure). The velocity that should be given to mass 2m (when rod is in horizontal position) in counter-clockwise direction so that the rod can just become vertical is :

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The correct option is C When the rod becomes vertical mv1 = 6mv2⇒v1 = 6v2 .........(i) 12 mv21 + 126mv22 = mgl .........(ii) Solve (i) and (ii) to get Vrel = v1 + v2 = √73gl

The correct option is C

When the rod becomes vertical

$m v_{1} = 6 m v_{2} \Rightarrow v_{1} = 6 v_{2}$ .........(i)

$\frac{1}{2} m v_{1}^{2} + \frac{1}{2} 6 m v_{2}^{2} = m g l$ .........(ii)

Solve (i) and (ii) to get $V_{r e l} = v_{1} + v_{2} = \sqrt{\frac{7}{3} g l}$