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PHXI05:LAWS OF MOTION

363502 A gun applies a force $F$ on a bullet which is given by $F = (100 - 0.5 \times 10^{5} t) N$ . The bullet emerges out with speed $400 m / s$ . Then, find out the impulse exerted till force on bullet becomes zero.

1

0.2 N - s

2

0.3 N - s

3

0.1 N - s

4

0.4 N - s

Explanation:

Force applied by gun,
$F = (100 - 0.5 \times 10^{5} t) N$
Speed of bullet, $v = 400 m / s$ ,
When $F = 0$ , then
$100 - 0.5 \times 10^{5} t = 0$
$\Rightarrow t = 2 \times 10^{- 3} s$
Impulse, $I = \int F d t$
$= \int_{0}^{2 \times 10^{- 3}} (100 - 0.5 \times 10^{5} t) d t$
$= {[100 t - 0.5 \times 10^{5} \cdot \frac{t^{2}}{2}]}^{2 \times 10^{- 3}}$ $= 100 \times 2 \times 10^{- 3} - \frac{0.5 \times 10^{5}}{2} \times 4 \times 10^{- 6} - 0$
$= 0.2 - 0.1 = 0.1 N - s$

AIIMS - 2019

PHXI05:LAWS OF MOTION

363503 A particle moves in the $x$ - $y$ plane under the influence of a force such that its linear momentum is $\vec{p} (t) = A [\hat{i} \cos (k t) - \hat{j} \sin (k t)]$ where $A$ and $k$ are constants. The angle between the force and momentum is

1

0^{\circ}

2

30^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

Given that $\vec{p} (t) = A (\hat{i} \cos k t - \hat{j} \sin k t)$
$\vec{F} = \frac{d}{d t} (\vec{p} (t)) = A k (- \hat{i} \sin k t - \hat{j} \cos k t)$
$\vec{F} \cdot \vec{p} = A^{2} k (- \cos k t \sin k t + \sin k t \cos k t) = 0$
$∴$ The momentum and force are perpendicular to each other at $90^{\circ}$ .

PHXI05:LAWS OF MOTION

363504 The position-time graph for two cars of the same mass is given. The ratio of momentum of the $car A$ and $car B$ is
supporting img

1

3 : 1

2

1 : 3

3

1 : \sqrt{2}

4

\sqrt{3} : 2

Explanation:

Angle of the line $A$ is $30^{\circ}$ and that of $B$ is $60^{\circ}$
$\begin{aligned} v_{B} = \tan 60^{\circ} = \sqrt{3} \\ v_{A} = \tan 30^{\circ} = \frac{1}{\sqrt{3}} \\ \frac{v_{A}}{v_{B}} = \frac{1}{3} \end{aligned}$

PHXI05:LAWS OF MOTION

363505 The area of $F$ - $t$ curve is $A$ , where ‘ $F$ ’ is the force acting on one mass due to the other. If one of the colliding bodies of mass $M$ is at rest initially, its speed just after the collision is:

1

\sqrt{\frac{2 A}{M}}

2

M / A

3

A M

4

A / M

Explanation:

The charge in momentum is equal to area of the $F - t$ curve. Here the mass is at rest initially
$M v = A \Rightarrow v = \frac{A}{M}$

PHXI05:LAWS OF MOTION

363502 A gun applies a force $F$ on a bullet which is given by $F = (100 - 0.5 \times 10^{5} t) N$ . The bullet emerges out with speed $400 m / s$ . Then, find out the impulse exerted till force on bullet becomes zero.

1

0.2 N - s

2

0.3 N - s

3

0.1 N - s

4

0.4 N - s

Explanation:

Force applied by gun,
$F = (100 - 0.5 \times 10^{5} t) N$
Speed of bullet, $v = 400 m / s$ ,
When $F = 0$ , then
$100 - 0.5 \times 10^{5} t = 0$
$\Rightarrow t = 2 \times 10^{- 3} s$
Impulse, $I = \int F d t$
$= \int_{0}^{2 \times 10^{- 3}} (100 - 0.5 \times 10^{5} t) d t$
$= {[100 t - 0.5 \times 10^{5} \cdot \frac{t^{2}}{2}]}^{2 \times 10^{- 3}}$ $= 100 \times 2 \times 10^{- 3} - \frac{0.5 \times 10^{5}}{2} \times 4 \times 10^{- 6} - 0$
$= 0.2 - 0.1 = 0.1 N - s$

AIIMS - 2019

PHXI05:LAWS OF MOTION

363503 A particle moves in the $x$ - $y$ plane under the influence of a force such that its linear momentum is $\vec{p} (t) = A [\hat{i} \cos (k t) - \hat{j} \sin (k t)]$ where $A$ and $k$ are constants. The angle between the force and momentum is

1

0^{\circ}

2

30^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

Given that $\vec{p} (t) = A (\hat{i} \cos k t - \hat{j} \sin k t)$
$\vec{F} = \frac{d}{d t} (\vec{p} (t)) = A k (- \hat{i} \sin k t - \hat{j} \cos k t)$
$\vec{F} \cdot \vec{p} = A^{2} k (- \cos k t \sin k t + \sin k t \cos k t) = 0$
$∴$ The momentum and force are perpendicular to each other at $90^{\circ}$ .

PHXI05:LAWS OF MOTION

363504 The position-time graph for two cars of the same mass is given. The ratio of momentum of the $car A$ and $car B$ is
supporting img

1

3 : 1

2

1 : 3

3

1 : \sqrt{2}

4

\sqrt{3} : 2

Explanation:

Angle of the line $A$ is $30^{\circ}$ and that of $B$ is $60^{\circ}$
$\begin{aligned} v_{B} = \tan 60^{\circ} = \sqrt{3} \\ v_{A} = \tan 30^{\circ} = \frac{1}{\sqrt{3}} \\ \frac{v_{A}}{v_{B}} = \frac{1}{3} \end{aligned}$

PHXI05:LAWS OF MOTION

363505 The area of $F$ - $t$ curve is $A$ , where ‘ $F$ ’ is the force acting on one mass due to the other. If one of the colliding bodies of mass $M$ is at rest initially, its speed just after the collision is:

1

\sqrt{\frac{2 A}{M}}

2

M / A

3

A M

4

A / M

Explanation:

The charge in momentum is equal to area of the $F - t$ curve. Here the mass is at rest initially
$M v = A \Rightarrow v = \frac{A}{M}$

PHXI05:LAWS OF MOTION

363502 A gun applies a force $F$ on a bullet which is given by $F = (100 - 0.5 \times 10^{5} t) N$ . The bullet emerges out with speed $400 m / s$ . Then, find out the impulse exerted till force on bullet becomes zero.

1

0.2 N - s

2

0.3 N - s

3

0.1 N - s

4

0.4 N - s

Explanation:

Force applied by gun,
$F = (100 - 0.5 \times 10^{5} t) N$
Speed of bullet, $v = 400 m / s$ ,
When $F = 0$ , then
$100 - 0.5 \times 10^{5} t = 0$
$\Rightarrow t = 2 \times 10^{- 3} s$
Impulse, $I = \int F d t$
$= \int_{0}^{2 \times 10^{- 3}} (100 - 0.5 \times 10^{5} t) d t$
$= {[100 t - 0.5 \times 10^{5} \cdot \frac{t^{2}}{2}]}^{2 \times 10^{- 3}}$ $= 100 \times 2 \times 10^{- 3} - \frac{0.5 \times 10^{5}}{2} \times 4 \times 10^{- 6} - 0$
$= 0.2 - 0.1 = 0.1 N - s$

AIIMS - 2019

PHXI05:LAWS OF MOTION

363503 A particle moves in the $x$ - $y$ plane under the influence of a force such that its linear momentum is $\vec{p} (t) = A [\hat{i} \cos (k t) - \hat{j} \sin (k t)]$ where $A$ and $k$ are constants. The angle between the force and momentum is

1

0^{\circ}

2

30^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

Given that $\vec{p} (t) = A (\hat{i} \cos k t - \hat{j} \sin k t)$
$\vec{F} = \frac{d}{d t} (\vec{p} (t)) = A k (- \hat{i} \sin k t - \hat{j} \cos k t)$
$\vec{F} \cdot \vec{p} = A^{2} k (- \cos k t \sin k t + \sin k t \cos k t) = 0$
$∴$ The momentum and force are perpendicular to each other at $90^{\circ}$ .

PHXI05:LAWS OF MOTION

363504 The position-time graph for two cars of the same mass is given. The ratio of momentum of the $car A$ and $car B$ is
supporting img

1

3 : 1

2

1 : 3

3

1 : \sqrt{2}

4

\sqrt{3} : 2

Explanation:

Angle of the line $A$ is $30^{\circ}$ and that of $B$ is $60^{\circ}$
$\begin{aligned} v_{B} = \tan 60^{\circ} = \sqrt{3} \\ v_{A} = \tan 30^{\circ} = \frac{1}{\sqrt{3}} \\ \frac{v_{A}}{v_{B}} = \frac{1}{3} \end{aligned}$

PHXI05:LAWS OF MOTION

363505 The area of $F$ - $t$ curve is $A$ , where ‘ $F$ ’ is the force acting on one mass due to the other. If one of the colliding bodies of mass $M$ is at rest initially, its speed just after the collision is:

1

\sqrt{\frac{2 A}{M}}

2

M / A

3

A M

4

A / M

Explanation:

The charge in momentum is equal to area of the $F - t$ curve. Here the mass is at rest initially
$M v = A \Rightarrow v = \frac{A}{M}$

PHXI05:LAWS OF MOTION

363502 A gun applies a force $F$ on a bullet which is given by $F = (100 - 0.5 \times 10^{5} t) N$ . The bullet emerges out with speed $400 m / s$ . Then, find out the impulse exerted till force on bullet becomes zero.

1

0.2 N - s

2

0.3 N - s

3

0.1 N - s

4

0.4 N - s

Explanation:

Force applied by gun,
$F = (100 - 0.5 \times 10^{5} t) N$
Speed of bullet, $v = 400 m / s$ ,
When $F = 0$ , then
$100 - 0.5 \times 10^{5} t = 0$
$\Rightarrow t = 2 \times 10^{- 3} s$
Impulse, $I = \int F d t$
$= \int_{0}^{2 \times 10^{- 3}} (100 - 0.5 \times 10^{5} t) d t$
$= {[100 t - 0.5 \times 10^{5} \cdot \frac{t^{2}}{2}]}^{2 \times 10^{- 3}}$ $= 100 \times 2 \times 10^{- 3} - \frac{0.5 \times 10^{5}}{2} \times 4 \times 10^{- 6} - 0$
$= 0.2 - 0.1 = 0.1 N - s$

AIIMS - 2019

PHXI05:LAWS OF MOTION

363503 A particle moves in the $x$ - $y$ plane under the influence of a force such that its linear momentum is $\vec{p} (t) = A [\hat{i} \cos (k t) - \hat{j} \sin (k t)]$ where $A$ and $k$ are constants. The angle between the force and momentum is

1

0^{\circ}

2

30^{\circ}

3

45^{\circ}

4

90^{\circ}

Explanation:

Given that $\vec{p} (t) = A (\hat{i} \cos k t - \hat{j} \sin k t)$
$\vec{F} = \frac{d}{d t} (\vec{p} (t)) = A k (- \hat{i} \sin k t - \hat{j} \cos k t)$
$\vec{F} \cdot \vec{p} = A^{2} k (- \cos k t \sin k t + \sin k t \cos k t) = 0$
$∴$ The momentum and force are perpendicular to each other at $90^{\circ}$ .

PHXI05:LAWS OF MOTION

363504 The position-time graph for two cars of the same mass is given. The ratio of momentum of the $car A$ and $car B$ is
supporting img

1

3 : 1

2

1 : 3

3

1 : \sqrt{2}

4

\sqrt{3} : 2

Explanation:

Angle of the line $A$ is $30^{\circ}$ and that of $B$ is $60^{\circ}$
$\begin{aligned} v_{B} = \tan 60^{\circ} = \sqrt{3} \\ v_{A} = \tan 30^{\circ} = \frac{1}{\sqrt{3}} \\ \frac{v_{A}}{v_{B}} = \frac{1}{3} \end{aligned}$

PHXI05:LAWS OF MOTION

363505 The area of $F$ - $t$ curve is $A$ , where ‘ $F$ ’ is the force acting on one mass due to the other. If one of the colliding bodies of mass $M$ is at rest initially, its speed just after the collision is:

1

\sqrt{\frac{2 A}{M}}

2

M / A

3

A M

4

A / M

Explanation:

The charge in momentum is equal to area of the $F - t$ curve. Here the mass is at rest initially
$M v = A \Rightarrow v = \frac{A}{M}$