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LAWS OF MOTION (ADDITIONAL)

371687 Three forces $A = (i + j + k), B = (2 i - j + 3 k)$ and $C$ acting on a body to keep it in equilibrium. Then $C$ is

1

- (3 i + 4 k)

2

- (4 i + 3 k)

3

3 i + 4 j

4

2 i - 3 k

Explanation:

A Given,
Three forces,
$A = (i + j + k)$
$B = (2 i - j + 3 k), C$
$∵ \vec{A} + \vec{B} + \vec{C} = 0$
$(\hat{i} + \hat{j} + \hat{k}) + (2 \hat{i} - \hat{j} + 3 \hat{k}) + \vec{C} = 0$
$3 \hat{i} + 4 \hat{k} + \vec{C} = 0$
$\vec{C} = - 3 \hat{i} - 4 \hat{j}$
$\vec{C} = - (3 \hat{i} + 4 \hat{j})$

AP EMCET(Medical)-2008

LAWS OF MOTION (ADDITIONAL)

371688 A mass $m$ falls freely from rest. The linear momentum, after it has fallen through a height $h$ , is :
( $g$ = acceleration due to gravity)

1

\sqrt{mgh}

2

m \sqrt{2 g h}

3

m \sqrt{gh}

4 zero

Explanation:

B We know that,
Newton's third's law of motion,
$v^{2} = u^{2} + 2 gh$
When initial velocity, $u = 0$
Then,
$v^{2} = 2 gh$
$v = \sqrt{2 gh}$
The linear momentum, $P = mv$
$P = m \times \sqrt{2 gh} {∵ v = \sqrt{2 gh}}$
$P = m \sqrt{2 gh}$

AP EAMCET(Medical)-1998

LAWS OF MOTION (ADDITIONAL)

371689 Two masses of $m$ and $4 m$ are moving with equal kinetic energy. The ratio of their linear momentum is

1

1 : 8

2

1 : 4

3

1 : 2

4

4 : 1

Explanation:

C Given,
$m_{1} = m$
$m_{2} = 4 m$
$(KE)_{1} = (KE)_{2}$
$P = \sqrt{2 m (KE)_{1}}$
$P_{2} = \sqrt{2 m_{2} (KE)_{2}}$
$P_{2} = \sqrt{2 \times 4 m (KE)_{2}}$
$\Rightarrow \frac{P_{1}}{P_{2}} = \sqrt{\frac{m}{4 m}} = \frac{1}{2} [∵ (KE)_{1} = (KE)_{2}]$

DCE-2009

LAWS OF MOTION (ADDITIONAL)

371690 The one which does not represent a force in any context is

1 friction

2 impulse

3 tension

4 weight

5 viscous drag

Explanation:

B Impulse is defined as overall effect of a force acting over time and it expressed in Newton-seconds. Impulse $=$ changes in Momentum $∴$ Impulse is does not represent force.

Kerala CEE - 2016

LAWS OF MOTION (ADDITIONAL)

371687 Three forces $A = (i + j + k), B = (2 i - j + 3 k)$ and $C$ acting on a body to keep it in equilibrium. Then $C$ is

1

- (3 i + 4 k)

2

- (4 i + 3 k)

3

3 i + 4 j

4

2 i - 3 k

Explanation:

A Given,
Three forces,
$A = (i + j + k)$
$B = (2 i - j + 3 k), C$
$∵ \vec{A} + \vec{B} + \vec{C} = 0$
$(\hat{i} + \hat{j} + \hat{k}) + (2 \hat{i} - \hat{j} + 3 \hat{k}) + \vec{C} = 0$
$3 \hat{i} + 4 \hat{k} + \vec{C} = 0$
$\vec{C} = - 3 \hat{i} - 4 \hat{j}$
$\vec{C} = - (3 \hat{i} + 4 \hat{j})$

AP EMCET(Medical)-2008

LAWS OF MOTION (ADDITIONAL)

371688 A mass $m$ falls freely from rest. The linear momentum, after it has fallen through a height $h$ , is :
( $g$ = acceleration due to gravity)

1

\sqrt{mgh}

2

m \sqrt{2 g h}

3

m \sqrt{gh}

4 zero

Explanation:

B We know that,
Newton's third's law of motion,
$v^{2} = u^{2} + 2 gh$
When initial velocity, $u = 0$
Then,
$v^{2} = 2 gh$
$v = \sqrt{2 gh}$
The linear momentum, $P = mv$
$P = m \times \sqrt{2 gh} {∵ v = \sqrt{2 gh}}$
$P = m \sqrt{2 gh}$

AP EAMCET(Medical)-1998

LAWS OF MOTION (ADDITIONAL)

371689 Two masses of $m$ and $4 m$ are moving with equal kinetic energy. The ratio of their linear momentum is

1

1 : 8

2

1 : 4

3

1 : 2

4

4 : 1

Explanation:

C Given,
$m_{1} = m$
$m_{2} = 4 m$
$(KE)_{1} = (KE)_{2}$
$P = \sqrt{2 m (KE)_{1}}$
$P_{2} = \sqrt{2 m_{2} (KE)_{2}}$
$P_{2} = \sqrt{2 \times 4 m (KE)_{2}}$
$\Rightarrow \frac{P_{1}}{P_{2}} = \sqrt{\frac{m}{4 m}} = \frac{1}{2} [∵ (KE)_{1} = (KE)_{2}]$

DCE-2009

LAWS OF MOTION (ADDITIONAL)

371690 The one which does not represent a force in any context is

1 friction

2 impulse

3 tension

4 weight

5 viscous drag

Explanation:

B Impulse is defined as overall effect of a force acting over time and it expressed in Newton-seconds. Impulse $=$ changes in Momentum $∴$ Impulse is does not represent force.

Kerala CEE - 2016

LAWS OF MOTION (ADDITIONAL)

371687 Three forces $A = (i + j + k), B = (2 i - j + 3 k)$ and $C$ acting on a body to keep it in equilibrium. Then $C$ is

1

- (3 i + 4 k)

2

- (4 i + 3 k)

3

3 i + 4 j

4

2 i - 3 k

Explanation:

A Given,
Three forces,
$A = (i + j + k)$
$B = (2 i - j + 3 k), C$
$∵ \vec{A} + \vec{B} + \vec{C} = 0$
$(\hat{i} + \hat{j} + \hat{k}) + (2 \hat{i} - \hat{j} + 3 \hat{k}) + \vec{C} = 0$
$3 \hat{i} + 4 \hat{k} + \vec{C} = 0$
$\vec{C} = - 3 \hat{i} - 4 \hat{j}$
$\vec{C} = - (3 \hat{i} + 4 \hat{j})$

AP EMCET(Medical)-2008

LAWS OF MOTION (ADDITIONAL)

371688 A mass $m$ falls freely from rest. The linear momentum, after it has fallen through a height $h$ , is :
( $g$ = acceleration due to gravity)

1

\sqrt{mgh}

2

m \sqrt{2 g h}

3

m \sqrt{gh}

4 zero

Explanation:

B We know that,
Newton's third's law of motion,
$v^{2} = u^{2} + 2 gh$
When initial velocity, $u = 0$
Then,
$v^{2} = 2 gh$
$v = \sqrt{2 gh}$
The linear momentum, $P = mv$
$P = m \times \sqrt{2 gh} {∵ v = \sqrt{2 gh}}$
$P = m \sqrt{2 gh}$

AP EAMCET(Medical)-1998

LAWS OF MOTION (ADDITIONAL)

371689 Two masses of $m$ and $4 m$ are moving with equal kinetic energy. The ratio of their linear momentum is

1

1 : 8

2

1 : 4

3

1 : 2

4

4 : 1

Explanation:

C Given,
$m_{1} = m$
$m_{2} = 4 m$
$(KE)_{1} = (KE)_{2}$
$P = \sqrt{2 m (KE)_{1}}$
$P_{2} = \sqrt{2 m_{2} (KE)_{2}}$
$P_{2} = \sqrt{2 \times 4 m (KE)_{2}}$
$\Rightarrow \frac{P_{1}}{P_{2}} = \sqrt{\frac{m}{4 m}} = \frac{1}{2} [∵ (KE)_{1} = (KE)_{2}]$

DCE-2009

LAWS OF MOTION (ADDITIONAL)

371690 The one which does not represent a force in any context is

1 friction

2 impulse

3 tension

4 weight

5 viscous drag

Explanation:

B Impulse is defined as overall effect of a force acting over time and it expressed in Newton-seconds. Impulse $=$ changes in Momentum $∴$ Impulse is does not represent force.

Kerala CEE - 2016

LAWS OF MOTION (ADDITIONAL)

371687 Three forces $A = (i + j + k), B = (2 i - j + 3 k)$ and $C$ acting on a body to keep it in equilibrium. Then $C$ is

1

- (3 i + 4 k)

2

- (4 i + 3 k)

3

3 i + 4 j

4

2 i - 3 k

Explanation:

A Given,
Three forces,
$A = (i + j + k)$
$B = (2 i - j + 3 k), C$
$∵ \vec{A} + \vec{B} + \vec{C} = 0$
$(\hat{i} + \hat{j} + \hat{k}) + (2 \hat{i} - \hat{j} + 3 \hat{k}) + \vec{C} = 0$
$3 \hat{i} + 4 \hat{k} + \vec{C} = 0$
$\vec{C} = - 3 \hat{i} - 4 \hat{j}$
$\vec{C} = - (3 \hat{i} + 4 \hat{j})$

AP EMCET(Medical)-2008

LAWS OF MOTION (ADDITIONAL)

371688 A mass $m$ falls freely from rest. The linear momentum, after it has fallen through a height $h$ , is :
( $g$ = acceleration due to gravity)

1

\sqrt{mgh}

2

m \sqrt{2 g h}

3

m \sqrt{gh}

4 zero

Explanation:

B We know that,
Newton's third's law of motion,
$v^{2} = u^{2} + 2 gh$
When initial velocity, $u = 0$
Then,
$v^{2} = 2 gh$
$v = \sqrt{2 gh}$
The linear momentum, $P = mv$
$P = m \times \sqrt{2 gh} {∵ v = \sqrt{2 gh}}$
$P = m \sqrt{2 gh}$

AP EAMCET(Medical)-1998

LAWS OF MOTION (ADDITIONAL)

371689 Two masses of $m$ and $4 m$ are moving with equal kinetic energy. The ratio of their linear momentum is

1

1 : 8

2

1 : 4

3

1 : 2

4

4 : 1

Explanation:

C Given,
$m_{1} = m$
$m_{2} = 4 m$
$(KE)_{1} = (KE)_{2}$
$P = \sqrt{2 m (KE)_{1}}$
$P_{2} = \sqrt{2 m_{2} (KE)_{2}}$
$P_{2} = \sqrt{2 \times 4 m (KE)_{2}}$
$\Rightarrow \frac{P_{1}}{P_{2}} = \sqrt{\frac{m}{4 m}} = \frac{1}{2} [∵ (KE)_{1} = (KE)_{2}]$

DCE-2009

LAWS OF MOTION (ADDITIONAL)

371690 The one which does not represent a force in any context is

1 friction

2 impulse

3 tension

4 weight

5 viscous drag

Explanation:

B Impulse is defined as overall effect of a force acting over time and it expressed in Newton-seconds. Impulse $=$ changes in Momentum $∴$ Impulse is does not represent force.

Kerala CEE - 2016

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