QBManager

LAWS OF MOTION (ADDITIONAL)

371657 A block $P$ of mass $M_{P}$ is in contact with another block $Q$ of mass $M_{Q}$ as shown in the figure and they are placed on a smooth floor. Force on block $Q$ is

1

\frac{M_{P}}{M_{P} + M_{Q}}

2

\frac{M_{Q} F}{M_{P} + M_{Q}}

3

\frac{M_{P} F}{M_{Q}}

4

\frac{M_{Q} F}{M_{P}}

Explanation:

B Given, mass of block $P = M_{P}$ , mass of block $Q = M_{Q}$

Since, $F_{ext} = ma$
$F = (M_{P} + M_{Q}) a$
$a = \frac{F}{M_{P} + M_{Q}}$
Now,

$∴ N = M_{Q} a$
Put the value of ' $a$ '
$N = \frac{M_{Q} F}{M_{P} + M_{Q}}$

VITEEE - 2017

LAWS OF MOTION (ADDITIONAL)

371658 The momentum of the particle at any instant is given by $3 \cos 4 t \hat{i} + 3 \sin 4 t \hat{j}$ . What is the angle between momentum and force acting on it?

1

60^{\circ}

2

30^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

D Given,
$\vec{p} = 3 \cos 4 t \hat{i} + 3 \sin 4 \hat{t}$
Force $= \frac{d \vec{p}}{dt} = \frac{d}{dt} (3 \cos 4 t \hat{i} + 3 \sin 4 \hat{j})$
$\vec{F} = - 12 \sin 4 t \hat{i} + 12 \cos 4 \hat{j}$
$\vec{F} \cdot \vec{P} = | \vec{F} | \cdot | \vec{P} | \cos θ$
$| \vec{F} | \cdot | \vec{P} | \cos θ = (- 12 \sin 4 t \hat{i} + 12 \cos 4 \hat{j}) (3 \cos 4 t \hat{i} + 3 \sin 4 \hat{j})$
$| \vec{F} | \cdot | \vec{P} | \cos θ = 36 (- \sin 4 t \cos 4 t + \cos 4 t \sin 4 t)$
$| \vec{F} | \cdot | \vec{P} | \cos θ = 36 [\sin (4 t - 4 t)]$
$| \vec{F} | \cdot | \vec{P} | \cos θ = 0$
$\cos θ = \cos 90^{\circ}$
$θ = 90^{\circ}$

J and K CET- 2005

LAWS OF MOTION (ADDITIONAL)

371659 A reference frame attached to earth cannot be an inertial frame because

1 Earth is revolving around the sun

2 Earth is rotating about its axis

3 Newton's laws are applicable in this frame

4 both (a) and (b)

JCECE-2007

LAWS OF MOTION (ADDITIONAL)

371660 A machine gun of mass $10 kg$ fires $30 g$ bullets at the rate of $6 bullet / s$ with a speed of $400 m / s$ . The force required to keep the gun in position will be

1

30 N

2

40 N

3

72 N

4

400 N

Explanation:

C Given that, $10 kg & 30 \times 10^{- 3} kg$ mass of machine gun and bullet respectively $& 6 bullet / s$ fired Let $v$ be the velocity of each unit,
According to conservation of momentum
$m_{1} \times v = m_{2} \times 400$
$10 \times v = 6 \times 30 \times 10^{- 3} \times 400$
$v = \frac{12 \times 6}{10} = 7.2 m / s$
$= \frac{Δ P}{Δ t} [∵ Δ t = 1 \sec]$
$= \frac{m_{1} v}{1}$
$= \frac{10 \times 7.2}{1} = 72 N$
$= 72 N$
$∴ v = \frac{12 \times 6}{10} = 7.2 m / s$

JCECE-2013

LAWS OF MOTION (ADDITIONAL)

371657 A block $P$ of mass $M_{P}$ is in contact with another block $Q$ of mass $M_{Q}$ as shown in the figure and they are placed on a smooth floor. Force on block $Q$ is

1

\frac{M_{P}}{M_{P} + M_{Q}}

2

\frac{M_{Q} F}{M_{P} + M_{Q}}

3

\frac{M_{P} F}{M_{Q}}

4

\frac{M_{Q} F}{M_{P}}

Explanation:

B Given, mass of block $P = M_{P}$ , mass of block $Q = M_{Q}$

Since, $F_{ext} = ma$
$F = (M_{P} + M_{Q}) a$
$a = \frac{F}{M_{P} + M_{Q}}$
Now,

$∴ N = M_{Q} a$
Put the value of ' $a$ '
$N = \frac{M_{Q} F}{M_{P} + M_{Q}}$

VITEEE - 2017

LAWS OF MOTION (ADDITIONAL)

371658 The momentum of the particle at any instant is given by $3 \cos 4 t \hat{i} + 3 \sin 4 t \hat{j}$ . What is the angle between momentum and force acting on it?

1

60^{\circ}

2

30^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

D Given,
$\vec{p} = 3 \cos 4 t \hat{i} + 3 \sin 4 \hat{t}$
Force $= \frac{d \vec{p}}{dt} = \frac{d}{dt} (3 \cos 4 t \hat{i} + 3 \sin 4 \hat{j})$
$\vec{F} = - 12 \sin 4 t \hat{i} + 12 \cos 4 \hat{j}$
$\vec{F} \cdot \vec{P} = | \vec{F} | \cdot | \vec{P} | \cos θ$
$| \vec{F} | \cdot | \vec{P} | \cos θ = (- 12 \sin 4 t \hat{i} + 12 \cos 4 \hat{j}) (3 \cos 4 t \hat{i} + 3 \sin 4 \hat{j})$
$| \vec{F} | \cdot | \vec{P} | \cos θ = 36 (- \sin 4 t \cos 4 t + \cos 4 t \sin 4 t)$
$| \vec{F} | \cdot | \vec{P} | \cos θ = 36 [\sin (4 t - 4 t)]$
$| \vec{F} | \cdot | \vec{P} | \cos θ = 0$
$\cos θ = \cos 90^{\circ}$
$θ = 90^{\circ}$

J and K CET- 2005

LAWS OF MOTION (ADDITIONAL)

371659 A reference frame attached to earth cannot be an inertial frame because

1 Earth is revolving around the sun

2 Earth is rotating about its axis

3 Newton's laws are applicable in this frame

4 both (a) and (b)

JCECE-2007

LAWS OF MOTION (ADDITIONAL)

371660 A machine gun of mass $10 kg$ fires $30 g$ bullets at the rate of $6 bullet / s$ with a speed of $400 m / s$ . The force required to keep the gun in position will be

1

30 N

2

40 N

3

72 N

4

400 N

Explanation:

C Given that, $10 kg & 30 \times 10^{- 3} kg$ mass of machine gun and bullet respectively $& 6 bullet / s$ fired Let $v$ be the velocity of each unit,
According to conservation of momentum
$m_{1} \times v = m_{2} \times 400$
$10 \times v = 6 \times 30 \times 10^{- 3} \times 400$
$v = \frac{12 \times 6}{10} = 7.2 m / s$
$= \frac{Δ P}{Δ t} [∵ Δ t = 1 \sec]$
$= \frac{m_{1} v}{1}$
$= \frac{10 \times 7.2}{1} = 72 N$
$= 72 N$
$∴ v = \frac{12 \times 6}{10} = 7.2 m / s$

JCECE-2013

LAWS OF MOTION (ADDITIONAL)

371657 A block $P$ of mass $M_{P}$ is in contact with another block $Q$ of mass $M_{Q}$ as shown in the figure and they are placed on a smooth floor. Force on block $Q$ is

1

\frac{M_{P}}{M_{P} + M_{Q}}

2

\frac{M_{Q} F}{M_{P} + M_{Q}}

3

\frac{M_{P} F}{M_{Q}}

4

\frac{M_{Q} F}{M_{P}}

Explanation:

B Given, mass of block $P = M_{P}$ , mass of block $Q = M_{Q}$

Since, $F_{ext} = ma$
$F = (M_{P} + M_{Q}) a$
$a = \frac{F}{M_{P} + M_{Q}}$
Now,

$∴ N = M_{Q} a$
Put the value of ' $a$ '
$N = \frac{M_{Q} F}{M_{P} + M_{Q}}$

VITEEE - 2017

LAWS OF MOTION (ADDITIONAL)

371658 The momentum of the particle at any instant is given by $3 \cos 4 t \hat{i} + 3 \sin 4 t \hat{j}$ . What is the angle between momentum and force acting on it?

1

60^{\circ}

2

30^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

D Given,
$\vec{p} = 3 \cos 4 t \hat{i} + 3 \sin 4 \hat{t}$
Force $= \frac{d \vec{p}}{dt} = \frac{d}{dt} (3 \cos 4 t \hat{i} + 3 \sin 4 \hat{j})$
$\vec{F} = - 12 \sin 4 t \hat{i} + 12 \cos 4 \hat{j}$
$\vec{F} \cdot \vec{P} = | \vec{F} | \cdot | \vec{P} | \cos θ$
$| \vec{F} | \cdot | \vec{P} | \cos θ = (- 12 \sin 4 t \hat{i} + 12 \cos 4 \hat{j}) (3 \cos 4 t \hat{i} + 3 \sin 4 \hat{j})$
$| \vec{F} | \cdot | \vec{P} | \cos θ = 36 (- \sin 4 t \cos 4 t + \cos 4 t \sin 4 t)$
$| \vec{F} | \cdot | \vec{P} | \cos θ = 36 [\sin (4 t - 4 t)]$
$| \vec{F} | \cdot | \vec{P} | \cos θ = 0$
$\cos θ = \cos 90^{\circ}$
$θ = 90^{\circ}$

J and K CET- 2005

LAWS OF MOTION (ADDITIONAL)

371659 A reference frame attached to earth cannot be an inertial frame because

1 Earth is revolving around the sun

2 Earth is rotating about its axis

3 Newton's laws are applicable in this frame

4 both (a) and (b)

JCECE-2007

LAWS OF MOTION (ADDITIONAL)

371660 A machine gun of mass $10 kg$ fires $30 g$ bullets at the rate of $6 bullet / s$ with a speed of $400 m / s$ . The force required to keep the gun in position will be

1

30 N

2

40 N

3

72 N

4

400 N

Explanation:

C Given that, $10 kg & 30 \times 10^{- 3} kg$ mass of machine gun and bullet respectively $& 6 bullet / s$ fired Let $v$ be the velocity of each unit,
According to conservation of momentum
$m_{1} \times v = m_{2} \times 400$
$10 \times v = 6 \times 30 \times 10^{- 3} \times 400$
$v = \frac{12 \times 6}{10} = 7.2 m / s$
$= \frac{Δ P}{Δ t} [∵ Δ t = 1 \sec]$
$= \frac{m_{1} v}{1}$
$= \frac{10 \times 7.2}{1} = 72 N$
$= 72 N$
$∴ v = \frac{12 \times 6}{10} = 7.2 m / s$

JCECE-2013

LAWS OF MOTION (ADDITIONAL)

371657 A block $P$ of mass $M_{P}$ is in contact with another block $Q$ of mass $M_{Q}$ as shown in the figure and they are placed on a smooth floor. Force on block $Q$ is

1

\frac{M_{P}}{M_{P} + M_{Q}}

2

\frac{M_{Q} F}{M_{P} + M_{Q}}

3

\frac{M_{P} F}{M_{Q}}

4

\frac{M_{Q} F}{M_{P}}

Explanation:

B Given, mass of block $P = M_{P}$ , mass of block $Q = M_{Q}$

Since, $F_{ext} = ma$
$F = (M_{P} + M_{Q}) a$
$a = \frac{F}{M_{P} + M_{Q}}$
Now,

$∴ N = M_{Q} a$
Put the value of ' $a$ '
$N = \frac{M_{Q} F}{M_{P} + M_{Q}}$

VITEEE - 2017

LAWS OF MOTION (ADDITIONAL)

371658 The momentum of the particle at any instant is given by $3 \cos 4 t \hat{i} + 3 \sin 4 t \hat{j}$ . What is the angle between momentum and force acting on it?

1

60^{\circ}

2

30^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

D Given,
$\vec{p} = 3 \cos 4 t \hat{i} + 3 \sin 4 \hat{t}$
Force $= \frac{d \vec{p}}{dt} = \frac{d}{dt} (3 \cos 4 t \hat{i} + 3 \sin 4 \hat{j})$
$\vec{F} = - 12 \sin 4 t \hat{i} + 12 \cos 4 \hat{j}$
$\vec{F} \cdot \vec{P} = | \vec{F} | \cdot | \vec{P} | \cos θ$
$| \vec{F} | \cdot | \vec{P} | \cos θ = (- 12 \sin 4 t \hat{i} + 12 \cos 4 \hat{j}) (3 \cos 4 t \hat{i} + 3 \sin 4 \hat{j})$
$| \vec{F} | \cdot | \vec{P} | \cos θ = 36 (- \sin 4 t \cos 4 t + \cos 4 t \sin 4 t)$
$| \vec{F} | \cdot | \vec{P} | \cos θ = 36 [\sin (4 t - 4 t)]$
$| \vec{F} | \cdot | \vec{P} | \cos θ = 0$
$\cos θ = \cos 90^{\circ}$
$θ = 90^{\circ}$

J and K CET- 2005

LAWS OF MOTION (ADDITIONAL)

371659 A reference frame attached to earth cannot be an inertial frame because

1 Earth is revolving around the sun

2 Earth is rotating about its axis

3 Newton's laws are applicable in this frame

4 both (a) and (b)

JCECE-2007

LAWS OF MOTION (ADDITIONAL)

371660 A machine gun of mass $10 kg$ fires $30 g$ bullets at the rate of $6 bullet / s$ with a speed of $400 m / s$ . The force required to keep the gun in position will be

1

30 N

2

40 N

3

72 N

4

400 N

Explanation:

C Given that, $10 kg & 30 \times 10^{- 3} kg$ mass of machine gun and bullet respectively $& 6 bullet / s$ fired Let $v$ be the velocity of each unit,
According to conservation of momentum
$m_{1} \times v = m_{2} \times 400$
$10 \times v = 6 \times 30 \times 10^{- 3} \times 400$
$v = \frac{12 \times 6}{10} = 7.2 m / s$
$= \frac{Δ P}{Δ t} [∵ Δ t = 1 \sec]$
$= \frac{m_{1} v}{1}$
$= \frac{10 \times 7.2}{1} = 72 N$
$= 72 N$
$∴ v = \frac{12 \times 6}{10} = 7.2 m / s$

JCECE-2013

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