Laws of Motion Test

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Laws of Motion Test - 35

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Question 1
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What is the value of $$p_1$$ and $$m_2$$ ?
m (kg) v (m/s) p = mv (kgm/s)
85 60 $$p_1$$
$$m_2$$ 2.5 6.25

A
5100, 2.5

B
5.1 , 2.5

C
2500, 5

D
55, 2.6

Solution

m (kg) v (m/s) p = mv (kgm/s)
85 60 5100
2.5
2.5 6.25
In first case:
$$ m=85kg $$
$$v =60 m/s$$
$$ p =mv = 85 \times 60 = 5100\ kg m/s$$
In second case:
$$ m=? $$
$$v =2.5 m/s$$
$$ p =mv = 6.25 $$
$$\therefore m = \dfrac{p}{v} = \dfrac{6.25}{2.5} = 2.5\ kg$$
Question 2
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A stone tied to a string of length $$L$$ is swired in a verticle circle with the other end of the string at the centre. At a certain instant of time the stone is at it lowest position and has a speed $$u$$. Find the magnitude of the change in its velocity as it reaches a position, where the string is horizontal.

A
$$\displaystyle \sqrt{(u^{2}-gL)}$$

B
$$\displaystyle \sqrt{2(u^{2}-gL)}$$

C
$$\displaystyle \sqrt{(u^{2}-2gL)}$$

D
$$\displaystyle \sqrt{2(u^{2}-2gL)}$$

Solution

By using conservation of energy,
$$\frac { 1 }{ 2 } m{ u }^{ 2 }=mgL+\frac { 1 }{ 2 } m{ v }^{ 2 }$$
$$\Rightarrow v=\sqrt { { u }^{ 2 }-2gL }$$
Change in velocity = $$\sqrt { { u }^{ 2 }-2gL } \hat { i } -u\hat { j }$$
Magnitude of change in velocity = $$\sqrt { { u }^{ 2 }-2gL+{ u }^{ 2 } } =\sqrt { 2\left( { u }^{ 2 }-gL \right) } $$
Question 3
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A bob is suspended from a crane by a cable of length $$5m$$. The crane and load are moving at a constant speed $$v_{0}$$. The crane is stopped by a bumper and the bob on the cable swings out an angle of $$60^{\circ}$$. Find the initial speed $$v_{0}.(g=9.8 m/s^{2})$$

A
$$2 ms^{-1}$$

B
$$3 ms^{-1}$$

C
$$5 ms^{-1}$$

D
$$7 ms^{-1}$$

Solution

By using conservation of energy, from the initial point to the highest point
$$\frac { 1 }{ 2 } m{ { v }_{ 0 } }^{ 2 }=mgl(1-cos{ 60 }^{ 0 })$$
$$\Rightarrow { v }_{ 0 }=\sqrt { 2gl(1-cos{ 60 }^{ 0 }) } =\sqrt { gl } =\sqrt { 50 } =7m/s$$
Question 4
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A simple pendulum is vibrating with angular amplitude of $$\theta=90^{o}$$ as shown in figure.
For what value of $$\theta$$ is the acceleration directed
$$(i)$$ Vertically upwards
$$(ii)$$ Horizontally
$$(iii)$$ Vertically downwards

A
$$0^{o},90^{o} \ cos^{-1}\displaystyle\frac{1}{\sqrt {3}},\ 90^{o}$$

B
$$ \cos^{-1}\displaystyle\frac{1}{\sqrt {3}},\ ,\ 0^{o},90^{o}$$

C
$$0^{o},90^{o}, \cos^{-1}\displaystyle\frac{1}{\sqrt{2}},$$

D
$$ \cos^{-1}\displaystyle\frac{1}{\sqrt {3}},\ 90^{o},\ 0^{o}$$

Solution

Acceleration of a simple pendulum is $$a = -ω^{2}\times A\times cos (ωt + θ)$$, where ω is angular velocity, $$A$$ is amplitude, $$t$$ is time and θ is angle.
If θ is $$0$$, then $$a = -ω^{2}\times A\times cos (ωt)$$ so direction is vertically upwards, when θ is $$cos^{-1} (1/3^{1/2})$$ then direction is horizontally, when θ is $$90^o$$, then $$a = ω^{2}\times A\times sin (ωt)$$ so direction is vertically downwards.
Question 5
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A pendulum bob is raised to a height, $$h$$, and released from rest. At what height will it attain half of its maximum speed?

A
$$\displaystyle\frac{3h}{4}$$

B
$$\displaystyle\frac{h}{2}$$

C
$$\displaystyle\frac{h}{4}$$

D
$$0.707h$$

Solution

By energy conservation, $$\dfrac{1}{2}mv^2_{max} = mgh$$
Let at height $$h'$$, it attain velocity $$v'=\dfrac{v_{max}}{2} = \sqrt{gh/2}$$
Now, $$mgh = \dfrac{1}{2}mv'^2 + mgh'$$

or $$mgh = \dfrac{mgh}{4} + mgh'$$

or $$mgh'=\dfrac{3}{4}mgh$$

$$\therefore h' = \dfrac{3h}{4}$$
Question 6
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Directions For Questions

A small particle of mass $$m$$ attached with a light inextensible thread of length $$L$$ is moving in a vertical circle. In the given case particle is moving in complete vertical circle and ratio of its maximum to minimum velocity is $$2:1$$.

...view full instructions

The kinetic energy of particle at the lower most position is

A
$$\displaystyle \frac{4mgL}{3}$$

B
$$2mgL$$

C
$$\displaystyle \frac{8mgL}{3}$$

D
$$\displaystyle \frac{2mgL}{3}$$

Solution

As $$v= 2\sqrt{\dfrac{gL}{3}}$$
Also, $$u= 2v \implies u= 4\sqrt{\dfrac{gL}{3}}$$
Thus $$K.E_ A= \dfrac{1}{2}mu^2= \dfrac{8mgL}{3}$$
Question 7
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Directions For Questions

Bob B of the pendulum AB is given an initial velocity $$\displaystyle \sqrt{3Lg}$$ in horizontal direction. Find the maximum height of the bob from the starting point:

...view full instructions

if AB is a massless string.

A
$$\displaystyle \frac{10L}{27}$$

B
$$\displaystyle \frac{20L}{27}$$

C
$$\displaystyle \frac{30L}{27}$$

D
$$\displaystyle \frac{40L}{27}$$

Solution

When the string has traversed an angle $$\theta$$ , then

$$T-mgcos\theta =\dfrac { m{ v }^{ 2 } }{ l }$$ , and conservation of energy gives

$$\dfrac { m{ u }^{ 2 } }{ 2 } =\dfrac { m{ v }^{ 2 } }{ 2 } +mgl(1-cos\theta ).$$ then the string becomes slacked tension in the string becomes zero. Solving the above two equations to get the value of $$\theta$$

$$\Rightarrow cos\theta =\dfrac { -1 }{ 3 } and { v }^{ 2 }={ u }^{ 2 }-2gl(1-cos\theta )=\dfrac { gl }{ 3 } $$ from that point the maximum height reached is $$\dfrac { { v }^{ 2 }{ sin }^{ 2 }\left( \pi -\theta \right) }{ 2g }$$ maximum height $$= l(1-cos\theta )+\dfrac { { v }^{ 2 }{ sin }^{ 2 }\left( \pi -\theta \right) }{ 2g } =\dfrac { 4l }{ 3 } +\dfrac { 4l }{ 27 } =\dfrac { 40l }{ 27 } $$
Question 8
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A simple pendulum of length $$l$$ has maximum angular displacement $$\displaystyle \theta .$$ Then maximum kinetic energy of a bob of mass $$m$$ is

A
$$\displaystyle \frac{1}{2}mgl$$

B
$$\displaystyle \frac{1}{2}mgl\cos \theta $$

C
$$\displaystyle mgl\left ( 1-\cos \theta \right )$$

D
$$\displaystyle \frac{1}{2}mgl\sin \theta $$

Solution

At point B, $$v=0$$ and $$P.E$$ is maximum.
At point A, $$P.E= 0$$ and $$K.E$$ is maximum.
Height of bob at B, $$AP= l-lcos\theta= l(1-cos\theta)$$
Thus $$P.E= mgl(1-cos\theta)$$
Applying conservation of energy at A and B, $$K.E_m +0 = 0 + P.E$$
$$\implies K.E_m=mgl (1-cos\theta) $$
Question 9
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A car is travelling along a circular curve that has a radius of $$50 m$$. If its speed is $$16 m/s$$ and is increasing uniformly at $$8 m/s$$ $$^{2}$$. Determine the magnitude of its acceleration at this instant.

A
5.5 m/s$$^{2}$$

B
7.5 m/s$$^{2}$$

C
9.5 m/s$$^{2}$$

D
12.5 m/s$$^{2}$$

Solution

Given: $$r =50 $$ m $$v = 16$$ m/s $$a_t =8 m/s^2$$
Radial acceleration of the car at that instant, $$a_r = \dfrac{v^2}{r} = \dfrac{16^2}{50} = 5.12$$ $$m/s^2$$

$$\therefore$$ Net acceleration of the car $$a = \sqrt{a_r^2 + a_t^2} = \sqrt{(5.12)^2 + 8^2} = 9.5 m/s^2$$
Question 10
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Directions For Questions

A small particle of mass $$m$$ attached with a light inextensible thread of length $$L$$ is moving in a vertical circle. In the given case particle is moving in complete vertical circle and ratio of its maximum to minimum velocity is $$2:1$$.

...view full instructions

Velocity of particle when it is moving vertically downward is

A
$$\displaystyle \sqrt{\frac{10gL}{3}}$$

B
$$\displaystyle 2\sqrt{\frac{gL}{3}}$$

C
$$\displaystyle \sqrt{\frac{8gL}{3}}$$

D
$$\displaystyle \sqrt{\frac{13gL}{3}}$$

Solution

Work-energy theorem for A to C, $$-mgL= \dfrac{1}{2}m(v')^2- \dfrac{1}{2}mu^2$$ where $$u= 4\sqrt{\dfrac{gL}{3}}$$
$$-mgL= \dfrac{1}{2}m(v')^2- \dfrac{8mgL}{3}$$
$$\implies v'= \sqrt{\dfrac{10 gL}{3}}$$