QBManager

LAWS OF MOTION (ADDITIONAL)

371744 At a metro station, a girl walks up a stationary escalator in $20 s$ . If she remains stationary on the escalator, then the escaltor take her up in 30s. The time taken by her to walk up on the moving escalator will be :

1

25 s

2

60 s

3

12 s

4

10 s

Explanation:

C Given,
Let distance $= h$
Walking time $= 20 \sec$
Escalator time $= 30 \sec$
Speed of $girl (v_{g}) = \frac{h}{20}$
Speed of escalator $(v_{e}) = \frac{h}{30}$
Time taken by girl when she walk upon the moving escalator is
$t = \frac{h}{v_{g} + v_{e}} \Rightarrow t = \frac{h}{\frac{h}{20} + \frac{h}{30}}$
$= \frac{20 \times 30}{(30 + 20)}$
$= 12 \sec$

Karnataka CET-2020

LAWS OF MOTION (ADDITIONAL)

371745 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:

C Given, $F = (100 - 0.5 \times 10^{5} t) N$
Impulse $= \int_{0}^{t} Fdt$
Now, time $t$ at which $F = 0$ is,
$0 = 100 - 0.5 \times 10^{5} t$
$t = 2 \times 10^{- 3}$
$∴$ Impulse $= \int_{0}^{2 \times 10^{- 3}} (100 - 0.5 \times 10^{5} t) dt$
$= {[100 t - \frac{0.5 \times 10^{5} t^{2}}{2}]}_{0}^{2 \times 10^{- 3}}$
$= [100 \times 2 \times 10^{- 3} - \frac{0.5 \times 10^{5} \times 4 \times 10^{- 6}}{2}]$
$= [200 \times 10^{- 3} - \frac{2}{2 \times 10}]$
$= [\frac{200}{1000} - \frac{1}{10}] = [\frac{2}{10} - \frac{1}{10}]$
$= \frac{1}{10}$
Impulse $= 0.1 N - s$ .

AIIMS-25.05.2019(M) Shift-1

LAWS OF MOTION (ADDITIONAL)

371746 The linear momentum of a particle moving in $X$ - $Y$ plane under the influence of a force is given as $\vec{P} (t) = A (\hat{i} \cos b t - \hat{j} \sin b t)$ where $A$ and $b$ are constant. The angle between the force and momentum is

1

0^{\circ}

2

45^{\circ}

3

60^{\circ}

4

90^{\circ}

Explanation:

D Given,
$\vec{P} = A (\hat{i} \cos b t - \hat{j} \sin b t)$
$\vec{F} = \frac{d \vec{p}}{d t} = \frac{d}{d t} A (\hat{i} \cos b t - \hat{j} \sin b t)$
$\vec{F} = A b [- \hat{i} \sin (b t) - \hat{j} \cos (b t)]$
Angle between two vector is $θ$
$\cos θ = \frac{F \cdot P}{| F | \cdot | P |}$
$\cos θ = \frac{A b (- \hat{i} \sin b t - \hat{j} \cos b t) A (\hat{i} \cos b t - \hat{j} \sin b t)}{| F | \cdot | P |}$
$\cos θ = \frac{A^{2} b (- \sin b t \cdot \cos b t + \sin b t \cdot \cos b t)}{| F | \cdot | P |}$
$∴ \cos θ = 0$
$\Rightarrow \cos θ = \cos 90^{\circ}$
$\Rightarrow θ = 90^{\circ}$

UPSEE 2019

LAWS OF MOTION (ADDITIONAL)

371747 Two objects of mass $m$ each moving with speed u m/s collide at $90^{\circ}$ , then final momentum is (assume collision is inelastic)

1

mu

2

2 mu

3

\sqrt{2} mu

4

2 \sqrt{2} mu

Explanation:

C Given,
Speed of objects $= u m / s$
Since, both object collide with $90^{\circ}$
According to law of conservation of momentum,
Total momentum before collision $=$ Total momentum after collision
$P_{i} = P_{f}$
$| m u \hat{i} + m u \hat{j} | = P_{f}$
$\sqrt{m^{2} u^{2} + m^{2} u^{2}} = P_{f}$
$P_{f} = \sqrt{2} m u$

JIPMER-2019

LAWS OF MOTION (ADDITIONAL)

371744 At a metro station, a girl walks up a stationary escalator in $20 s$ . If she remains stationary on the escalator, then the escaltor take her up in 30s. The time taken by her to walk up on the moving escalator will be :

1

25 s

2

60 s

3

12 s

4

10 s

Explanation:

C Given,
Let distance $= h$
Walking time $= 20 \sec$
Escalator time $= 30 \sec$
Speed of $girl (v_{g}) = \frac{h}{20}$
Speed of escalator $(v_{e}) = \frac{h}{30}$
Time taken by girl when she walk upon the moving escalator is
$t = \frac{h}{v_{g} + v_{e}} \Rightarrow t = \frac{h}{\frac{h}{20} + \frac{h}{30}}$
$= \frac{20 \times 30}{(30 + 20)}$
$= 12 \sec$

Karnataka CET-2020

LAWS OF MOTION (ADDITIONAL)

371745 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:

C Given, $F = (100 - 0.5 \times 10^{5} t) N$
Impulse $= \int_{0}^{t} Fdt$
Now, time $t$ at which $F = 0$ is,
$0 = 100 - 0.5 \times 10^{5} t$
$t = 2 \times 10^{- 3}$
$∴$ Impulse $= \int_{0}^{2 \times 10^{- 3}} (100 - 0.5 \times 10^{5} t) dt$
$= {[100 t - \frac{0.5 \times 10^{5} t^{2}}{2}]}_{0}^{2 \times 10^{- 3}}$
$= [100 \times 2 \times 10^{- 3} - \frac{0.5 \times 10^{5} \times 4 \times 10^{- 6}}{2}]$
$= [200 \times 10^{- 3} - \frac{2}{2 \times 10}]$
$= [\frac{200}{1000} - \frac{1}{10}] = [\frac{2}{10} - \frac{1}{10}]$
$= \frac{1}{10}$
Impulse $= 0.1 N - s$ .

AIIMS-25.05.2019(M) Shift-1

LAWS OF MOTION (ADDITIONAL)

371746 The linear momentum of a particle moving in $X$ - $Y$ plane under the influence of a force is given as $\vec{P} (t) = A (\hat{i} \cos b t - \hat{j} \sin b t)$ where $A$ and $b$ are constant. The angle between the force and momentum is

1

0^{\circ}

2

45^{\circ}

3

60^{\circ}

4

90^{\circ}

Explanation:

D Given,
$\vec{P} = A (\hat{i} \cos b t - \hat{j} \sin b t)$
$\vec{F} = \frac{d \vec{p}}{d t} = \frac{d}{d t} A (\hat{i} \cos b t - \hat{j} \sin b t)$
$\vec{F} = A b [- \hat{i} \sin (b t) - \hat{j} \cos (b t)]$
Angle between two vector is $θ$
$\cos θ = \frac{F \cdot P}{| F | \cdot | P |}$
$\cos θ = \frac{A b (- \hat{i} \sin b t - \hat{j} \cos b t) A (\hat{i} \cos b t - \hat{j} \sin b t)}{| F | \cdot | P |}$
$\cos θ = \frac{A^{2} b (- \sin b t \cdot \cos b t + \sin b t \cdot \cos b t)}{| F | \cdot | P |}$
$∴ \cos θ = 0$
$\Rightarrow \cos θ = \cos 90^{\circ}$
$\Rightarrow θ = 90^{\circ}$

UPSEE 2019

LAWS OF MOTION (ADDITIONAL)

371747 Two objects of mass $m$ each moving with speed u m/s collide at $90^{\circ}$ , then final momentum is (assume collision is inelastic)

1

mu

2

2 mu

3

\sqrt{2} mu

4

2 \sqrt{2} mu

Explanation:

C Given,
Speed of objects $= u m / s$
Since, both object collide with $90^{\circ}$
According to law of conservation of momentum,
Total momentum before collision $=$ Total momentum after collision
$P_{i} = P_{f}$
$| m u \hat{i} + m u \hat{j} | = P_{f}$
$\sqrt{m^{2} u^{2} + m^{2} u^{2}} = P_{f}$
$P_{f} = \sqrt{2} m u$

JIPMER-2019

LAWS OF MOTION (ADDITIONAL)

371744 At a metro station, a girl walks up a stationary escalator in $20 s$ . If she remains stationary on the escalator, then the escaltor take her up in 30s. The time taken by her to walk up on the moving escalator will be :

1

25 s

2

60 s

3

12 s

4

10 s

Explanation:

C Given,
Let distance $= h$
Walking time $= 20 \sec$
Escalator time $= 30 \sec$
Speed of $girl (v_{g}) = \frac{h}{20}$
Speed of escalator $(v_{e}) = \frac{h}{30}$
Time taken by girl when she walk upon the moving escalator is
$t = \frac{h}{v_{g} + v_{e}} \Rightarrow t = \frac{h}{\frac{h}{20} + \frac{h}{30}}$
$= \frac{20 \times 30}{(30 + 20)}$
$= 12 \sec$

Karnataka CET-2020

LAWS OF MOTION (ADDITIONAL)

371745 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:

C Given, $F = (100 - 0.5 \times 10^{5} t) N$
Impulse $= \int_{0}^{t} Fdt$
Now, time $t$ at which $F = 0$ is,
$0 = 100 - 0.5 \times 10^{5} t$
$t = 2 \times 10^{- 3}$
$∴$ Impulse $= \int_{0}^{2 \times 10^{- 3}} (100 - 0.5 \times 10^{5} t) dt$
$= {[100 t - \frac{0.5 \times 10^{5} t^{2}}{2}]}_{0}^{2 \times 10^{- 3}}$
$= [100 \times 2 \times 10^{- 3} - \frac{0.5 \times 10^{5} \times 4 \times 10^{- 6}}{2}]$
$= [200 \times 10^{- 3} - \frac{2}{2 \times 10}]$
$= [\frac{200}{1000} - \frac{1}{10}] = [\frac{2}{10} - \frac{1}{10}]$
$= \frac{1}{10}$
Impulse $= 0.1 N - s$ .

AIIMS-25.05.2019(M) Shift-1

LAWS OF MOTION (ADDITIONAL)

371746 The linear momentum of a particle moving in $X$ - $Y$ plane under the influence of a force is given as $\vec{P} (t) = A (\hat{i} \cos b t - \hat{j} \sin b t)$ where $A$ and $b$ are constant. The angle between the force and momentum is

1

0^{\circ}

2

45^{\circ}

3

60^{\circ}

4

90^{\circ}

Explanation:

D Given,
$\vec{P} = A (\hat{i} \cos b t - \hat{j} \sin b t)$
$\vec{F} = \frac{d \vec{p}}{d t} = \frac{d}{d t} A (\hat{i} \cos b t - \hat{j} \sin b t)$
$\vec{F} = A b [- \hat{i} \sin (b t) - \hat{j} \cos (b t)]$
Angle between two vector is $θ$
$\cos θ = \frac{F \cdot P}{| F | \cdot | P |}$
$\cos θ = \frac{A b (- \hat{i} \sin b t - \hat{j} \cos b t) A (\hat{i} \cos b t - \hat{j} \sin b t)}{| F | \cdot | P |}$
$\cos θ = \frac{A^{2} b (- \sin b t \cdot \cos b t + \sin b t \cdot \cos b t)}{| F | \cdot | P |}$
$∴ \cos θ = 0$
$\Rightarrow \cos θ = \cos 90^{\circ}$
$\Rightarrow θ = 90^{\circ}$

UPSEE 2019

LAWS OF MOTION (ADDITIONAL)

371747 Two objects of mass $m$ each moving with speed u m/s collide at $90^{\circ}$ , then final momentum is (assume collision is inelastic)

1

mu

2

2 mu

3

\sqrt{2} mu

4

2 \sqrt{2} mu

Explanation:

C Given,
Speed of objects $= u m / s$
Since, both object collide with $90^{\circ}$
According to law of conservation of momentum,
Total momentum before collision $=$ Total momentum after collision
$P_{i} = P_{f}$
$| m u \hat{i} + m u \hat{j} | = P_{f}$
$\sqrt{m^{2} u^{2} + m^{2} u^{2}} = P_{f}$
$P_{f} = \sqrt{2} m u$

JIPMER-2019

LAWS OF MOTION (ADDITIONAL)

371744 At a metro station, a girl walks up a stationary escalator in $20 s$ . If she remains stationary on the escalator, then the escaltor take her up in 30s. The time taken by her to walk up on the moving escalator will be :

1

25 s

2

60 s

3

12 s

4

10 s

Explanation:

C Given,
Let distance $= h$
Walking time $= 20 \sec$
Escalator time $= 30 \sec$
Speed of $girl (v_{g}) = \frac{h}{20}$
Speed of escalator $(v_{e}) = \frac{h}{30}$
Time taken by girl when she walk upon the moving escalator is
$t = \frac{h}{v_{g} + v_{e}} \Rightarrow t = \frac{h}{\frac{h}{20} + \frac{h}{30}}$
$= \frac{20 \times 30}{(30 + 20)}$
$= 12 \sec$

Karnataka CET-2020

LAWS OF MOTION (ADDITIONAL)

371745 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:

C Given, $F = (100 - 0.5 \times 10^{5} t) N$
Impulse $= \int_{0}^{t} Fdt$
Now, time $t$ at which $F = 0$ is,
$0 = 100 - 0.5 \times 10^{5} t$
$t = 2 \times 10^{- 3}$
$∴$ Impulse $= \int_{0}^{2 \times 10^{- 3}} (100 - 0.5 \times 10^{5} t) dt$
$= {[100 t - \frac{0.5 \times 10^{5} t^{2}}{2}]}_{0}^{2 \times 10^{- 3}}$
$= [100 \times 2 \times 10^{- 3} - \frac{0.5 \times 10^{5} \times 4 \times 10^{- 6}}{2}]$
$= [200 \times 10^{- 3} - \frac{2}{2 \times 10}]$
$= [\frac{200}{1000} - \frac{1}{10}] = [\frac{2}{10} - \frac{1}{10}]$
$= \frac{1}{10}$
Impulse $= 0.1 N - s$ .

AIIMS-25.05.2019(M) Shift-1

LAWS OF MOTION (ADDITIONAL)

371746 The linear momentum of a particle moving in $X$ - $Y$ plane under the influence of a force is given as $\vec{P} (t) = A (\hat{i} \cos b t - \hat{j} \sin b t)$ where $A$ and $b$ are constant. The angle between the force and momentum is

1

0^{\circ}

2

45^{\circ}

3

60^{\circ}

4

90^{\circ}

Explanation:

D Given,
$\vec{P} = A (\hat{i} \cos b t - \hat{j} \sin b t)$
$\vec{F} = \frac{d \vec{p}}{d t} = \frac{d}{d t} A (\hat{i} \cos b t - \hat{j} \sin b t)$
$\vec{F} = A b [- \hat{i} \sin (b t) - \hat{j} \cos (b t)]$
Angle between two vector is $θ$
$\cos θ = \frac{F \cdot P}{| F | \cdot | P |}$
$\cos θ = \frac{A b (- \hat{i} \sin b t - \hat{j} \cos b t) A (\hat{i} \cos b t - \hat{j} \sin b t)}{| F | \cdot | P |}$
$\cos θ = \frac{A^{2} b (- \sin b t \cdot \cos b t + \sin b t \cdot \cos b t)}{| F | \cdot | P |}$
$∴ \cos θ = 0$
$\Rightarrow \cos θ = \cos 90^{\circ}$
$\Rightarrow θ = 90^{\circ}$

UPSEE 2019

LAWS OF MOTION (ADDITIONAL)

371747 Two objects of mass $m$ each moving with speed u m/s collide at $90^{\circ}$ , then final momentum is (assume collision is inelastic)

1

mu

2

2 mu

3

\sqrt{2} mu

4

2 \sqrt{2} mu

Explanation:

C Given,
Speed of objects $= u m / s$
Since, both object collide with $90^{\circ}$
According to law of conservation of momentum,
Total momentum before collision $=$ Total momentum after collision
$P_{i} = P_{f}$
$| m u \hat{i} + m u \hat{j} | = P_{f}$
$\sqrt{m^{2} u^{2} + m^{2} u^{2}} = P_{f}$
$P_{f} = \sqrt{2} m u$

JIPMER-2019

Question Bank Series

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