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Rotational Motion

269597 Four identical planks each of lengths ' $L$ ' are arranged one above the other over a table as shown. Each projects a distance 'a' beyond the edge of the one that is below it. What is the maximum possible value of ' $a$ ' for the system to be in equilibrium without tripping forward?

1

L / 5

2

L / 4

3

L / 3

4

L

Explanation:

CM of bricks, above each brick must not be beyond its edge. $x_{cm} = \frac{Σ m_{i} x_{i}}{Σ m_{i}}; x_{c m} = L$
$x_{1} = a + \frac{L}{2}, x_{2} = 2 a + \frac{L}{2}, x_{3} = 3 a + \frac{L}{2}$
(or) $a = \frac{L}{n}$

Rotational Motion

269598 Two masses ' $m_{1}$ ' and ' $m_{2}$ ' $(m_{1} < m_{2})$ are connected to the ends of a light inextensible string which passes over the surface of a smooth fixed pulley. If the system is released from rest, the acceleration of the centre of mass of the system will be ( $g =$ acceleration due to gravity)

1

\frac{g (m_{1} - m_{2})}{(m_{1} + m_{2})}

2

\frac{g {(m_{1} - m_{2})}^{2}}{{(m_{1} + m_{2})}^{2}}

3

\frac{g (m_{1} + m_{2})}{(m_{1} - m_{2})}

4

\frac{g (m_{1} + m_{2})}{(m_{1} - m_{2})}

Explanation:

${(a_{c m})}_{y} = \frac{F_{e x t}}{M} = \frac{(m_{1} + m_{2}) g - 2 T}{m_{1} + m_{2}} \to (1)$
But $T = \frac{2 m_{1} m_{2} g}{m_{1} + m_{2}} \to (2)$

Rotational Motion

269599 Two bodies of masses $m_{1}$ and $m_{2}$ are moving with velocity $v_{1}$ and $v_{2}$ respectively in the same direction. The total momentum of the system in the frame of reference attached to the centre of mass is ( $v$ is relative velocity between the masses)

1

\frac{m_{1} m_{2} v}{m_{1} - m_{2}}

2

\frac{2 m_{1} m_{2} v}{m_{1} + m_{2}}

3 zero

4

\frac{4 m_{1} m_{2} v}{m_{1} + m_{2}}

Rotational Motion

269600 A shell in flight explodes into $n$ equal fragments $k$ of the fragments reach the ground earlier than the other fragments. The acceleration of their centre of mass subsequently will be

1

g

2

n - k) g

3

\frac{(n - k) g}{k}

4

\frac{(n - k)}{n} g

Explanation:

$a_{c m} = \frac{\sum m_{i} a_{i}}{\sum m_{i}}$

Rotational Motion

269597 Four identical planks each of lengths ' $L$ ' are arranged one above the other over a table as shown. Each projects a distance 'a' beyond the edge of the one that is below it. What is the maximum possible value of ' $a$ ' for the system to be in equilibrium without tripping forward?

1

L / 5

2

L / 4

3

L / 3

4

L

Explanation:

CM of bricks, above each brick must not be beyond its edge. $x_{cm} = \frac{Σ m_{i} x_{i}}{Σ m_{i}}; x_{c m} = L$
$x_{1} = a + \frac{L}{2}, x_{2} = 2 a + \frac{L}{2}, x_{3} = 3 a + \frac{L}{2}$
(or) $a = \frac{L}{n}$

Rotational Motion

269598 Two masses ' $m_{1}$ ' and ' $m_{2}$ ' $(m_{1} < m_{2})$ are connected to the ends of a light inextensible string which passes over the surface of a smooth fixed pulley. If the system is released from rest, the acceleration of the centre of mass of the system will be ( $g =$ acceleration due to gravity)

1

\frac{g (m_{1} - m_{2})}{(m_{1} + m_{2})}

2

\frac{g {(m_{1} - m_{2})}^{2}}{{(m_{1} + m_{2})}^{2}}

3

\frac{g (m_{1} + m_{2})}{(m_{1} - m_{2})}

4

\frac{g (m_{1} + m_{2})}{(m_{1} - m_{2})}

Explanation:

${(a_{c m})}_{y} = \frac{F_{e x t}}{M} = \frac{(m_{1} + m_{2}) g - 2 T}{m_{1} + m_{2}} \to (1)$
But $T = \frac{2 m_{1} m_{2} g}{m_{1} + m_{2}} \to (2)$

Rotational Motion

269599 Two bodies of masses $m_{1}$ and $m_{2}$ are moving with velocity $v_{1}$ and $v_{2}$ respectively in the same direction. The total momentum of the system in the frame of reference attached to the centre of mass is ( $v$ is relative velocity between the masses)

1

\frac{m_{1} m_{2} v}{m_{1} - m_{2}}

2

\frac{2 m_{1} m_{2} v}{m_{1} + m_{2}}

3 zero

4

\frac{4 m_{1} m_{2} v}{m_{1} + m_{2}}

Rotational Motion

269600 A shell in flight explodes into $n$ equal fragments $k$ of the fragments reach the ground earlier than the other fragments. The acceleration of their centre of mass subsequently will be

1

g

2

n - k) g

3

\frac{(n - k) g}{k}

4

\frac{(n - k)}{n} g

Explanation:

$a_{c m} = \frac{\sum m_{i} a_{i}}{\sum m_{i}}$

Rotational Motion

269597 Four identical planks each of lengths ' $L$ ' are arranged one above the other over a table as shown. Each projects a distance 'a' beyond the edge of the one that is below it. What is the maximum possible value of ' $a$ ' for the system to be in equilibrium without tripping forward?

1

L / 5

2

L / 4

3

L / 3

4

L

Explanation:

CM of bricks, above each brick must not be beyond its edge. $x_{cm} = \frac{Σ m_{i} x_{i}}{Σ m_{i}}; x_{c m} = L$
$x_{1} = a + \frac{L}{2}, x_{2} = 2 a + \frac{L}{2}, x_{3} = 3 a + \frac{L}{2}$
(or) $a = \frac{L}{n}$

Rotational Motion

269598 Two masses ' $m_{1}$ ' and ' $m_{2}$ ' $(m_{1} < m_{2})$ are connected to the ends of a light inextensible string which passes over the surface of a smooth fixed pulley. If the system is released from rest, the acceleration of the centre of mass of the system will be ( $g =$ acceleration due to gravity)

1

\frac{g (m_{1} - m_{2})}{(m_{1} + m_{2})}

2

\frac{g {(m_{1} - m_{2})}^{2}}{{(m_{1} + m_{2})}^{2}}

3

\frac{g (m_{1} + m_{2})}{(m_{1} - m_{2})}

4

\frac{g (m_{1} + m_{2})}{(m_{1} - m_{2})}

Explanation:

${(a_{c m})}_{y} = \frac{F_{e x t}}{M} = \frac{(m_{1} + m_{2}) g - 2 T}{m_{1} + m_{2}} \to (1)$
But $T = \frac{2 m_{1} m_{2} g}{m_{1} + m_{2}} \to (2)$

Rotational Motion

269599 Two bodies of masses $m_{1}$ and $m_{2}$ are moving with velocity $v_{1}$ and $v_{2}$ respectively in the same direction. The total momentum of the system in the frame of reference attached to the centre of mass is ( $v$ is relative velocity between the masses)

1

\frac{m_{1} m_{2} v}{m_{1} - m_{2}}

2

\frac{2 m_{1} m_{2} v}{m_{1} + m_{2}}

3 zero

4

\frac{4 m_{1} m_{2} v}{m_{1} + m_{2}}

Rotational Motion

269600 A shell in flight explodes into $n$ equal fragments $k$ of the fragments reach the ground earlier than the other fragments. The acceleration of their centre of mass subsequently will be

1

g

2

n - k) g

3

\frac{(n - k) g}{k}

4

\frac{(n - k)}{n} g

Explanation:

$a_{c m} = \frac{\sum m_{i} a_{i}}{\sum m_{i}}$

Rotational Motion

269597 Four identical planks each of lengths ' $L$ ' are arranged one above the other over a table as shown. Each projects a distance 'a' beyond the edge of the one that is below it. What is the maximum possible value of ' $a$ ' for the system to be in equilibrium without tripping forward?

1

L / 5

2

L / 4

3

L / 3

4

L

Explanation:

CM of bricks, above each brick must not be beyond its edge. $x_{cm} = \frac{Σ m_{i} x_{i}}{Σ m_{i}}; x_{c m} = L$
$x_{1} = a + \frac{L}{2}, x_{2} = 2 a + \frac{L}{2}, x_{3} = 3 a + \frac{L}{2}$
(or) $a = \frac{L}{n}$

Rotational Motion

269598 Two masses ' $m_{1}$ ' and ' $m_{2}$ ' $(m_{1} < m_{2})$ are connected to the ends of a light inextensible string which passes over the surface of a smooth fixed pulley. If the system is released from rest, the acceleration of the centre of mass of the system will be ( $g =$ acceleration due to gravity)

1

\frac{g (m_{1} - m_{2})}{(m_{1} + m_{2})}

2

\frac{g {(m_{1} - m_{2})}^{2}}{{(m_{1} + m_{2})}^{2}}

3

\frac{g (m_{1} + m_{2})}{(m_{1} - m_{2})}

4

\frac{g (m_{1} + m_{2})}{(m_{1} - m_{2})}

Explanation:

${(a_{c m})}_{y} = \frac{F_{e x t}}{M} = \frac{(m_{1} + m_{2}) g - 2 T}{m_{1} + m_{2}} \to (1)$
But $T = \frac{2 m_{1} m_{2} g}{m_{1} + m_{2}} \to (2)$

Rotational Motion

269599 Two bodies of masses $m_{1}$ and $m_{2}$ are moving with velocity $v_{1}$ and $v_{2}$ respectively in the same direction. The total momentum of the system in the frame of reference attached to the centre of mass is ( $v$ is relative velocity between the masses)

1

\frac{m_{1} m_{2} v}{m_{1} - m_{2}}

2

\frac{2 m_{1} m_{2} v}{m_{1} + m_{2}}

3 zero

4

\frac{4 m_{1} m_{2} v}{m_{1} + m_{2}}

Rotational Motion

269600 A shell in flight explodes into $n$ equal fragments $k$ of the fragments reach the ground earlier than the other fragments. The acceleration of their centre of mass subsequently will be

1

g

2

n - k) g

3

\frac{(n - k) g}{k}

4

\frac{(n - k)}{n} g

Explanation:

$a_{c m} = \frac{\sum m_{i} a_{i}}{\sum m_{i}}$