Laws of Motion Test

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Laws of Motion Test - 66

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Question 1
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Directions For Questions

A small sphere B of mass $$m$$ is released from rest in the position shown and swings freely in a vertical plane, first about O and then about the peg A after the cord comes in contact with the peg. Determine the tension in the cord:

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just before the sphere comes in contact with the peg.

A
$$\displaystyle \frac{mg}{2}$$

B
$$\displaystyle {mg}$$

C
$$\displaystyle \frac{3mg}{2}$$

D
$$\displaystyle \frac{5mg}{2}$$

Solution

Given : $$r = 0.8$$ m $$u = 0$$ m/s
Before coming in contact with the peg, the sphere rotates in a circle of radius $$r =0.8$$ m
Let the velocity of the sphere when it reaches the point C be $$v$$.
From figure, $$PC = r sin 30^o = \dfrac{r}{2}$$ m

Work-energy theorem : $$W = \Delta K.E$$
$$\therefore$$ $$mg (PC) =\dfrac{1}{2} mv^2 - \dfrac{1}{2}mu^2 $$

OR $$mg \times \dfrac{r}{2} = \dfrac{mv^2}{2} - 0$$

$$\implies \dfrac{mv^2}{r} =mg$$

Using: $$ ma_r = \dfrac{mv^2}{r} = T - mgsin30^o$$
$$\therefore$$ $$mg = T - \dfrac{mg}{2}$$
$$\implies T = \dfrac{3mg}{2}$$
Question 2
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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:

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if AB is a massless rod,

A
$$\displaystyle \frac{L}{2}$$

B
$$\displaystyle \frac{3L}{2}$$

C
$$\displaystyle \frac{5L}{2}$$

D
$$\displaystyle \frac{7L}{2}$$

Solution

Initial velocity of the rod $$u = \sqrt{3Lg}$$
Let the maximum height attained to be $$H$$. At the starting point, its potential energy is zero while at the highest point, its kinetic is zero.
Applying conservation of energy : $$P.E_i + K.E_ i =P.E_f + K.E_f$$
$$\therefore$$ $$0 + \dfrac{1}{2}mu^2 = mgH + 0$$

$$\implies$$ $$H = \dfrac{u^2}{2g} = \dfrac{3Lg}{2g} = \dfrac{3L}{2}$$
Question 3
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A particle is given an initial speed $$u$$ inside a smooth spherical shell of radius $$R$$ so that it is just able to complete the circle. Acceleration of the particle, when its velocity is vertical, is

A
$$\displaystyle g\sqrt{10}$$

B
$$\displaystyle g$$

C
$$\displaystyle g\sqrt{2}$$

D
$$\displaystyle g\sqrt{6}$$

Solution

Let, the velocity at highest point be v then
$$m{ v }^{ 2 }/r = mg \dfrac { m{ u }^{ 2 } }{ 2 } = 2\times mgR + \dfrac { mv^{ 2 } }{ 2 } $$
$$ \Rightarrow u = \sqrt { 5Rg } \dfrac { mx^{ 2 } }{ 2 } = \dfrac { m{ u }^{ 2 } }{ 2 } - mgR$$
Net acceleration at the point, when velocity is vertical is:
$$a = \sqrt { { \left( \dfrac { mx^{ 2 } }{ 2 } \right) }^{ 2 }+ { g }^{ 2 } } = \sqrt { 10\times g } $$
$$=g\sqrt{10}$$
Question 4
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A simple pendulum is released from rest with the string in horizontal position. The vertical component of the velocity of the bob becomes maximum, when the string makes an angle $$\displaystyle \theta $$ with the vertical. The angle $$\displaystyle \theta $$ is equal to

A
$$\displaystyle \frac{\pi }{4}$$

B
$$\displaystyle \cos ^{-1}\left ( \frac{1}{\sqrt{3}} \right )$$

C
$$\displaystyle \sin ^{-1}\left ( \frac{1}{\sqrt{3}} \right )$$

D
$$\displaystyle \frac{\pi }{3}$$

Solution

At angle theta , l = length of rod v = pendulum velocity u = vertical velocity
By energy conservation,
$$mgl\cos { \theta } = \dfrac { m{ v }^{ 2 } }{ 2 } = \dfrac { m{ u }^{ 2 } }{ 2\sin ^{ 2 }{ \theta } } $$
Differentiating wrt $$\theta$$ , and putting $$\dfrac { du }{ d\theta } =0$$;
we get, $$\theta = \cos ^{ -1 }{ \left( \dfrac { 1 }{ \sqrt { 3 } } \right) } \\ $$
Question 5
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A particle is moving in a circular path in the vertical plane. It is attached at one end of a string of length $$l$$ whose other end is fixed. The velocity at lowest point is $$u$$. The tension in the string is $$\displaystyle \vec{T}$$ and acceleration of the particle is $$\displaystyle \vec{a}$$ at any position. Then $$\displaystyle \vec{T}.\vec{a}$$ is zero at highest point if

A
$$\displaystyle u> \sqrt{5gl}$$

B
$$\displaystyle u= \sqrt{5gl}$$

C
Both (a) and (b) correct

D
Both (a) and (b) are wrong

Solution

Tension is $$0$$ in case of $$B$$.
If $$u$$ is more than this, then Tension is never zero.
if $$u$$ is less than this, then the particle would never be able to complete the circular loop.
$$m{ v }^{ 2 }/r\quad =\quad mg\\ \dfrac { m{ u }^{ 2 } }{ 2 } =\quad 2\times mgR\quad +\quad \dfrac { mv^{ 2 } }{ 2 } \\ \Rightarrow \quad u\quad =\quad \sqrt { 5Rg } $$
Question 6
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What is the maximum speed at which a railway carriage can move without toppling over along a curve of radius $$R=200m$$ if the distance from the centre of gravity of the carriage to the level of the rails is $$h=1.0m$$, the distance between the rails is $$h=1.0m$$, the distance between the rails is $$l=2.0m$$ and the rails are laid horizontally? (Take $$\displaystyle g= 10m/s^{2}$$)

A
$$\displaystyle 11.18ms^{-1}$$

B
$$\displaystyle 22.36ms^{-1}$$

C
$$\displaystyle 44.72ms^{-1}$$

D
$$\displaystyle 74ms^{-1}$$

Solution

In railway carriage's frame, the carriage is in translational as well as rotational equilibrium.
Rotational equilibrium : Torque about point O is zero i.e $$\tau_{o} = 0$$
$$\therefore$$ $$f \times 1 - N \times 1 = 0$$ $$\implies f = N$$
Translational equilibrium :
$$\therefore$$ $$mg = N$$ $$\implies mg = f$$ ..........(1)
Also $$f = ma_r$$ .........(2)
From (1) and (2), we get $$ma_r = mg$$
$$\therefore$$ $$\dfrac{mv^2}{r} = mg$$ $$\implies v = \sqrt{rg}$$
$$\therefore$$ $$v = \sqrt{200 \times 10} =44.72 $$ $$ m/s$$
Question 7
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The sphere at A is given a downward velocity $$\displaystyle v_{0}$$ of magnitude $$5 m/s$$ and swings in a vertical plane at the end of a rope of length $$l=2 m$$ attached to a support at $$O$$. Determine the angle $$\displaystyle \theta $$ at which the rope will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere.

A
$$\displaystyle \sin ^{-1}\left ( \frac{1}{4} \right )$$

B
$$\displaystyle \sin ^{-1}\left ( \frac{1}{3} \right )$$

C
$$\displaystyle \sin ^{-1}\left ( \frac{1}{2} \right )$$

D
$$\displaystyle \sin ^{-1}\left ( \frac{3}{4} \right )$$

Solution

Given : $$v_o =5 m/s$$ $$l =2$$ m $$T = 2mg$$
From figure, $$PM = l sin\theta$$
Circular motion equation at P : $$\dfrac{mv^2}{l} = T - mgsin\theta$$
OR $$mv^2 =l( 2mg - mgsin\theta) = 2 (2 m \times 10 - m(10) sin\theta)$$
$$\implies $$ $$mv^2 = m (40 - 20 sin\theta)$$

Work-energy theorem : $$W = \Delta K.E$$
$$\therefore$$ $$mg (PM) = \dfrac{1}{2}mv^2 - \dfrac{1}{2}mv_o^2$$

OR $$2mg (l sin\theta) =m(40 - 20sin\theta) - mv_o^2$$
OR $$2 m (10) (2 sin\theta) = m (40 - 20 sin\theta) - m(25)$$ $$\implies sin\theta = \dfrac{1}{4}$$
Question 8
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A weightless thread can withstand tension upto $$30\;N$$. A stone of mass $$0.5\ kg$$ is tied to it and is revolved in a circular path of radius $$2\ m$$ in a vertical plane. If $$g=10\ m/s^2$$, then the maximum angular velocity of the stone will be

A
$$5\ rad/s$$

B
$$\sqrt{30}\ rad/s$$

C
$$\sqrt{60}\ rad/s$$

D
$$10\ rad/s$$

Solution

Given : $$T_{max} =30$$ N $$m = 0.5 kg$$ $$r =2$$ m $$g = 10 m/s^2$$
Tension in the string is maximum when the stone is at the lowest position of the vertical motion.
Let the maximum angular velocity be $$w_{max}$$.
Using circular motion equation : $$mrw^2_{max} =T_{max} - mg$$
$$\therefore$$ $$0.5 \times 2 \times w^2_{max} =30 - 0.5 \times 10$$
OR $$w^2_{max} = 30 - 5 =25$$ $$\implies w_{max} = 5$$ $$rad/s$$
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 $$\displaystyle 8 m/s^{2},$$ determine the magnitude of its acceleration at this instant.

A
$$9.5\ m/s^2$$

B
$$9.8\ m/s^2$$

C
$$8.5\ m/s^2$$

D
$$7.5\ m/s^2$$

Solution

$$\displaystyle a= \sqrt{{a_{t}}^{2}+{a_{n}}^{2}}= \sqrt{\left ( \frac{dv}{dt} \right )^{2}+\left ( \frac{v^{2}}{R} \right )^{2}}= \sqrt{\left ( 8 \right )^{2}+\left ( \frac{256}{50} \right )^{2}}= 9.5m/s^{2}$$
Question 10
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A circular tube of mass M is placed vertically on a horizontal surface as shown in the figure. Two small spheres, each of mass m, just fit in the tube, are released from the top. If $$\displaystyle \theta $$ gives the angle between radius vector of either ball with the vertical, obtain the value of the ratio M/m if the tube breaks its contact with ground when $$\displaystyle \theta = 60^{\circ}.$$ Neglect any friction.

A
$$1:2$$

B
$$1:3$$

C
$$2:1$$

D
$$2:3$$

Solution

Let the speed of sphere at $$\theta=60^0$$ be $$v$$.
By conservation of energy, $$\dfrac{1}{2}mv^2=mg(R-Rcos60^0)=mgR/2$$

$$\implies mv^2=mgR$$

Let the normal force by tube on a sphere be $$N$$, towards the center.
Balancing forces in the radial direction,
$$N+mgcos60^0=\dfrac{mv^2}{R}=\dfrac{mgR}{R}=mg$$
$$\implies N=0.5mg -(1)$$

Now, by newton's third law, equal normal force is applied by each sphere on the tube in the radial direction. Total vertical component of these forces is,
$$N_{vertical}=2Ncos60^0=N$$, in the upward direction

The tube will break contact with ground when normal force by ground on the tube is zero.

So, $$N_{vertical}=Mg\implies 0.5mg=Mg\implies M=0.5m$$
$$\implies M:m=0.5:1=1:2$$