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PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365715 A pulley fixed to the ceiling carries a string with blocks of mass $m$ and $3 m$ attached to its ends. The masses of string and pulley are negligible. When the system is released,its center of mass moves with acceleration of

1

g / 4

2 0

3

g / 6

4

g / 2

Explanation:

When the system is released, heavier mass move downward and the lighter one upward.
Acceleration of the blocks is
supporting img

$\begin{aligned} a = (\frac{3 m - m}{3 m + m}) g = \frac{g}{2} \\ {\vec{a}}_{c m} = \frac{m_{1} {\vec{a}}_{1} + m_{2} {\vec{a}}_{2}}{m_{1} + m_{2}} = \frac{m a \hat{j} - 3 m a \hat{j}}{4 m} \\ {\vec{a}}_{c m} = - \frac{a}{2} \hat{j} = \frac{- g}{4} \hat{j} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365716 Both the blocks shown in the given arrangement are given together a horizontal velocity towards right. If $x {m s}^{- 2}$ be the subsequent acceleration of centre of mass of the system of blocks, find the minimum value of $x$
(Given, $m_{A} = 2 m_{B}$ and $g = 10 {m s}^{- 2}$ , all in SI units)
supporting img

1 5

2 2

3 7

4 8

Explanation:

$F_{ext} \geq (0.2) (2 m + m) (10) = 6 m$
$\begin{aligned} {\bar{a}}_{C O M} = \frac{F_{e x I}}{2 m + m} = \frac{6 m}{2 m + m} = 2 {m s}^{- 2} \\ ∴ x = 2 \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365717 When a force is applied on a body, Newton's second law is applicable to

1 Centre of mass

2 Any part of the body

3 Upper most part of body

4 Lower most part of body

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365718 Two particles of mass $5 k g$ and $10 k g$ respectively are attached to the two ends of a rigid rod of length $1 m$ with negligible mass. The distance of centre of mass from $5 k g$ mass is

1

50 c m

2

67 c m

3

80 c m

4

33 c m

Explanation:

supporting img
$\frac{r_{1}}{r_{2}} = \frac{m_{2}}{m_{1}} = \frac{10}{5} = \frac{2}{1}$
$r_{1} = \frac{2}{3} r = \frac{2}{3} \times 1 m = 67 c m$

NEET - 2020

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365715 A pulley fixed to the ceiling carries a string with blocks of mass $m$ and $3 m$ attached to its ends. The masses of string and pulley are negligible. When the system is released,its center of mass moves with acceleration of

1

g / 4

2 0

3

g / 6

4

g / 2

Explanation:

When the system is released, heavier mass move downward and the lighter one upward.
Acceleration of the blocks is
supporting img

$\begin{aligned} a = (\frac{3 m - m}{3 m + m}) g = \frac{g}{2} \\ {\vec{a}}_{c m} = \frac{m_{1} {\vec{a}}_{1} + m_{2} {\vec{a}}_{2}}{m_{1} + m_{2}} = \frac{m a \hat{j} - 3 m a \hat{j}}{4 m} \\ {\vec{a}}_{c m} = - \frac{a}{2} \hat{j} = \frac{- g}{4} \hat{j} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365716 Both the blocks shown in the given arrangement are given together a horizontal velocity towards right. If $x {m s}^{- 2}$ be the subsequent acceleration of centre of mass of the system of blocks, find the minimum value of $x$
(Given, $m_{A} = 2 m_{B}$ and $g = 10 {m s}^{- 2}$ , all in SI units)
supporting img

1 5

2 2

3 7

4 8

Explanation:

$F_{ext} \geq (0.2) (2 m + m) (10) = 6 m$
$\begin{aligned} {\bar{a}}_{C O M} = \frac{F_{e x I}}{2 m + m} = \frac{6 m}{2 m + m} = 2 {m s}^{- 2} \\ ∴ x = 2 \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365717 When a force is applied on a body, Newton's second law is applicable to

1 Centre of mass

2 Any part of the body

3 Upper most part of body

4 Lower most part of body

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365718 Two particles of mass $5 k g$ and $10 k g$ respectively are attached to the two ends of a rigid rod of length $1 m$ with negligible mass. The distance of centre of mass from $5 k g$ mass is

1

50 c m

2

67 c m

3

80 c m

4

33 c m

Explanation:

supporting img
$\frac{r_{1}}{r_{2}} = \frac{m_{2}}{m_{1}} = \frac{10}{5} = \frac{2}{1}$
$r_{1} = \frac{2}{3} r = \frac{2}{3} \times 1 m = 67 c m$

NEET - 2020

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365715 A pulley fixed to the ceiling carries a string with blocks of mass $m$ and $3 m$ attached to its ends. The masses of string and pulley are negligible. When the system is released,its center of mass moves with acceleration of

1

g / 4

2 0

3

g / 6

4

g / 2

Explanation:

When the system is released, heavier mass move downward and the lighter one upward.
Acceleration of the blocks is
supporting img

$\begin{aligned} a = (\frac{3 m - m}{3 m + m}) g = \frac{g}{2} \\ {\vec{a}}_{c m} = \frac{m_{1} {\vec{a}}_{1} + m_{2} {\vec{a}}_{2}}{m_{1} + m_{2}} = \frac{m a \hat{j} - 3 m a \hat{j}}{4 m} \\ {\vec{a}}_{c m} = - \frac{a}{2} \hat{j} = \frac{- g}{4} \hat{j} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365716 Both the blocks shown in the given arrangement are given together a horizontal velocity towards right. If $x {m s}^{- 2}$ be the subsequent acceleration of centre of mass of the system of blocks, find the minimum value of $x$
(Given, $m_{A} = 2 m_{B}$ and $g = 10 {m s}^{- 2}$ , all in SI units)
supporting img

1 5

2 2

3 7

4 8

Explanation:

$F_{ext} \geq (0.2) (2 m + m) (10) = 6 m$
$\begin{aligned} {\bar{a}}_{C O M} = \frac{F_{e x I}}{2 m + m} = \frac{6 m}{2 m + m} = 2 {m s}^{- 2} \\ ∴ x = 2 \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365717 When a force is applied on a body, Newton's second law is applicable to

1 Centre of mass

2 Any part of the body

3 Upper most part of body

4 Lower most part of body

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365718 Two particles of mass $5 k g$ and $10 k g$ respectively are attached to the two ends of a rigid rod of length $1 m$ with negligible mass. The distance of centre of mass from $5 k g$ mass is

1

50 c m

2

67 c m

3

80 c m

4

33 c m

Explanation:

supporting img
$\frac{r_{1}}{r_{2}} = \frac{m_{2}}{m_{1}} = \frac{10}{5} = \frac{2}{1}$
$r_{1} = \frac{2}{3} r = \frac{2}{3} \times 1 m = 67 c m$

NEET - 2020

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365715 A pulley fixed to the ceiling carries a string with blocks of mass $m$ and $3 m$ attached to its ends. The masses of string and pulley are negligible. When the system is released,its center of mass moves with acceleration of

1

g / 4

2 0

3

g / 6

4

g / 2

Explanation:

When the system is released, heavier mass move downward and the lighter one upward.
Acceleration of the blocks is
supporting img

$\begin{aligned} a = (\frac{3 m - m}{3 m + m}) g = \frac{g}{2} \\ {\vec{a}}_{c m} = \frac{m_{1} {\vec{a}}_{1} + m_{2} {\vec{a}}_{2}}{m_{1} + m_{2}} = \frac{m a \hat{j} - 3 m a \hat{j}}{4 m} \\ {\vec{a}}_{c m} = - \frac{a}{2} \hat{j} = \frac{- g}{4} \hat{j} \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365716 Both the blocks shown in the given arrangement are given together a horizontal velocity towards right. If $x {m s}^{- 2}$ be the subsequent acceleration of centre of mass of the system of blocks, find the minimum value of $x$
(Given, $m_{A} = 2 m_{B}$ and $g = 10 {m s}^{- 2}$ , all in SI units)
supporting img

1 5

2 2

3 7

4 8

Explanation:

$F_{ext} \geq (0.2) (2 m + m) (10) = 6 m$
$\begin{aligned} {\bar{a}}_{C O M} = \frac{F_{e x I}}{2 m + m} = \frac{6 m}{2 m + m} = 2 {m s}^{- 2} \\ ∴ x = 2 \end{aligned}$

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365717 When a force is applied on a body, Newton's second law is applicable to

1 Centre of mass

2 Any part of the body

3 Upper most part of body

4 Lower most part of body

PHXI07:SYSTEMS OF PARTICLES AND ROTATIONAL MOTION

365718 Two particles of mass $5 k g$ and $10 k g$ respectively are attached to the two ends of a rigid rod of length $1 m$ with negligible mass. The distance of centre of mass from $5 k g$ mass is

1

50 c m

2

67 c m

3

80 c m

4

33 c m

Explanation:

supporting img
$\frac{r_{1}}{r_{2}} = \frac{m_{2}}{m_{1}} = \frac{10}{5} = \frac{2}{1}$
$r_{1} = \frac{2}{3} r = \frac{2}{3} \times 1 m = 67 c m$

NEET - 2020

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