Work Energy and Power Test

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Work Energy and Power Test - 20

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Weekly Quiz Competition

Question 1
1 / -0

The length of a ballistic pendulum is $$1 m$$ and mass of its block is $$1.9 kg$$. A bullet of mass $$0.1 kg$$ strikes the block of ballistic pendulum in horizontal direction with a velocity $$100ms^{–1}$$ and got embedded in the block. After collision the combined mass (block & bullet) swings away from lowest point. The tension in the string when it makes an angle $$60°$$ with vertical is $$(g=10ms^{-2})$$:

A
$$20\ N$$

B
$$30\ N$$

C
$$40\ N$$

D
$$50\ N$$

Solution

Momentum Conservation
$$0.1\times 100= \left ( 1.9+0.1 \right )V_{o}$$
$$V_o = 5m s^{-1}.$$
$$ H= L(1-cos60^{\circ}). =\dfrac{1}{2}= \dfrac{1}{2}m.$$

Energy Conservation
$$\dfrac{1}{2}(2)(5)^{2} = \dfrac{1}{2} \times 2 \times V^{2}+2\times10 \times \dfrac{1}{2}$$
$$25-10= V^{2}.$$
$$ V=\sqrt{15}mvs $$

By Force Balance
$$ T-mg\ cos\ 60^{\circ}= m\dfrac{V^{2}}{L}.$$
$$ T= 2\times10\times\dfrac{1}{2}+2\times\dfrac{15}{1}$$
$$= 40N.$$
Question 2
1 / -0

A simple pendulum is oscillating with an angular amplitude $$60^{0}$$. If mass of bob is $$50\ gram$$, the tension in the string at mean position is:
$$(g=10\ m/s^{2})$$

A
$$0.5\ N$$

B
$$1\ N$$

C
$$1.5\ N$$

D
$$2\ N$$

Solution

$$H = R(1 - cos\ 60^0)$$

$$H = \dfrac {R}{2}$$

Apply Work Energy Theorem,

$$mg\dfrac {R}{2} = \dfrac {1}{2}mV^2$$

$$V = \sqrt {Rg}$$

At mean position,
$$T - mg = \dfrac {mV^2}{R}$$

$$T = 2\ mg$$

$$= 2 \times 0.05 \times 10$$
$$= 1\ N$$
Question 3
1 / -0

A water bucket of mass '$$m$$' is revolved in a vertical circle with the help of a rope of length '$$r$$'. If the velocity of the bucket at the lowest point is $$\sqrt{7gr}$$ . Then the velocity and tension in the rope at the highest point are:

A
$$\sqrt{3gr},2mg$$

B
$$\sqrt{2gr},mg$$

C
$$\sqrt{gr},mg$$

D
$$Zero$$ , $$Zero$$

Solution

$$V_{b} = \sqrt{7gr}$$
Work Energy Theorem applied between top and bottom.
$$mg(2R) = \dfrac{1}{2}m (V_{b} )^{2} - \dfrac{1}{2}m (V_{T} )^{2}$$
$$2mgr = \dfrac{7}{2}mgr - \dfrac{1}{2}m (V_{T} )^{2}$$
$$V_{T} = \sqrt{3gr}$$

At top point:
$$T = \dfrac{mV_{T}^{2}}{r} - mg$$
$$=m \dfrac{3gr}{r} -mg$$
$$= 2mg$$
Question 4
1 / -0

The bob of a simple pendulum of mass $$'m'$$ is performing oscillations such that the tension in the string is equal to twice the weight of the bob while it is crossing the mean position. The tension in the string when the bob reaches extreme position is :

A
$$\dfrac{mg}{2}$$

B
$$mg$$

C
$$\dfrac{3mg}{2}$$

D
zero

Solution

At bottom $$T = 2mg$$
$$T - mg = \dfrac{mV^2}{R}$$
$$2mg - mg = \dfrac{mV^2}{R}$$
$$\sqrt {Rg} = V$$
Applying Work Energy Theorem between pt. (1) and (2)
$$mg R(1-cos\theta) = \dfrac{1}{2}mV^2$$
$$mg R(1-cos\theta) = \dfrac{mRg}{2}$$

here, $$\theta$$ is the angle made by the particle with the axis at its extreme point,

$$1-cos\theta = \dfrac{1}{2}$$
$$\theta = 60^0$$
at, $$\theta = 60^0$$
$$ T = mgcos\theta$$
$$\therefore T = mg\dfrac{1}{2}$$
Question 5
1 / -0

A simple pendulum consists of a light string from which a spherical bob of mass, $$M$$, is suspended. The distance between the point of suspension and the center of bob is $$L$$. At the lowest position, the bob is given tangential velocity of $$\sqrt{5gL}$$. The K.E of the bob when the string becomes horizontal is:

A
$$0$$

B
$$\dfrac{MgL}{2}$$

C
$$\dfrac{3MgL}{2}$$

D
$$\dfrac{5MgL}{2}$$

Solution

From Work Energy Theorem,

$$(WD)_G = K_f - K_i$$ , where $$(WD)_G$$ is work done by gravity

$$-MgL = \dfrac{-1}{2}M\times 5gL + K_i$$

$$K_i = \dfrac {3}{2}MgL$$
Question 6
1 / -0

Two bodies A, B of masses $$m_1, m_2$$ are knotted to a mass-less string at different points rotated along concentric circles in horizontal plane. The distances of A, B from common centre are 50cm, 1m. If the tensions in the string between centre to A and A to B are in the ratio 5:4, then the ratio of $$m_{1}$$ to $$m_{2}$$ is:

A
2 : 3

B
3 : 2

C
1 : 1

D
1 : 2

Solution

The tension will be more between center to A as it has to bear the centripetal force of 2 bodies.

applying force balance on the masses,
$$T_{1}-T_2=m_1 \omega^{2}r_1$$.........(1)
$$T_2=m_2 \omega^{2}r_2$$.........(1)

$$\Rightarrow \dfrac { { T }_{ 2 } }{ { T }_{ 1 } } =\dfrac { 4 }{ 5 } \quad $$

Now: $${ r }_{ 1 } = 0.5$$, $${ r }_{ 2 } =1 $$

=> $$\dfrac { 5 }{ 4 } = \dfrac { { m }_{ 1 }(0.5) }{ { m }_{ 2 }+1 } $$

=> $$\dfrac{m_1}{m_2} = \dfrac{2\times 1}{4} = \dfrac{1}{2}$$
Question 7
1 / -0

The bob of a simple pendulum is given a velocity in horizontal direction when the bob is at lowest position ,so that the bob describes vertical circle of radius equal to length of pendulum and tension in the string is $$10 \ N$$ when the bob is at an angle $$60^0$$ from lowest position of vertical circle. The tension in the string when the bob reaches highest position is (The mass of bob is $$ 100$$ gram. $$g = 10\ ms^{–2}$$):

A
$$9\ N$$

B
$$7 \ N$$

C
$$5.5\ N$$

D
$$3.5 \ N$$

Solution

For a body of mass $$'m'$$ moving in a vertical circle, Tension in the string at highest point and Tension at any angle $$\theta$$ with vertical is related as:-
$$T-{ T }_{ H }=3mg(1+cos\theta )$$
Substituting the values
$$10-{ T }_{ H }=3(0.1)(10)(1+cos60^0)=4.5\\ \Rightarrow { T }_{ H }=5.5\ N$$
Question 8
1 / -0

lf the two bodies moving at right angles collide and their initial momenta are $$\vec{\mathrm{P}}_{1}$$ and $$\vec{\mathrm{P}}_{2}$$, their resultant momentum after collison is :

A
$$\vec{\mathrm{P}_{1}}-\vec{\mathrm{P}}_{2}$$

B
$$\vec{\mathrm{P}}_{1}\sim\vec{\mathrm{P}}_{2}$$

C
$$\sqrt{\vec{\mathrm{P}_{1}}^{2}+\vec{\mathrm{P}}_{2}^{2}}$$

D
$$\sqrt{\vec{\mathrm{P}_{1}}^{2}-\vec{\mathrm{P}}_{2}^{2}}$$

Solution

By momentum conservation,
Resultant momentum after collision = Resultant momentum before collision = $$\sqrt{\vec {P_1}^2 + \vec {P_2}^2}$$
Question 9
1 / -0

The length of a simple pendulum is '$$L$$'. Its bob from rest position is projected horizontally with a velocity $$\sqrt{\dfrac{7gL}{2}}$$. The maximum angular displacement of bob, such that the string does not slack, is:

A
$$30^o$$

B
$$60^o$$

C
$$120^o$$

D
$$150^o$$

Solution

At point where tension become just zero, the centripetal force must be balanced by gravitational force
$$ mg cos\theta = \dfrac{mv_{1}^{2}}{L}$$ ------ (1)
BY conservation of energy
$$\dfrac{1}{2}m \dfrac{7gL}{2} = \dfrac{mv_{1}^{2}}{2}+mgL(1+cos\theta)$$
$$\sqrt{2gL} \sqrt{\dfrac{3}{4}-cos\theta} = v_{1}$$
Put $$v_{1}$$ in (1)
$$mg$$$$ cos\theta = $$$$\dfrac{m\times2gL}{L}\left ( \dfrac{3}{4}-cos\theta \right ).$$
$$cos\theta =\dfrac{3}{2}- 2cos\theta$$
$$cos\theta=\dfrac{1}{2}$$
$$\theta=60^{0}$$
$$\therefore $$Total maximum angular displacement
$$= 180^0-60^0$$
$$=120^{0}$$
Question 10
1 / -0

Instantaneous power of constant force acting on a particle moving in a straight line under the action of this force

A
is constant

B
increases linearly with time

C
decreases linearly with time

D
either increases or decreases linearly with time

Solution

Power is defined as rate of work done
$$\implies$$Power, $$P = F.v$$
Even though the force is constant, due to the force the velocity increases because of which its power increases.
Also from equations of motion we know that $$v = u + at$$
$$\implies$$ instantaneous power increases linearly with time

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Re-Attempt Weekly Quiz Competition

Work Energy and Power Test - 20

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Work Energy and Power Test - 20

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SHARING IS CARING

Question 1

$$20\ N$$

$$30\ N$$

$$40\ N$$

$$50\ N$$

Solution

Question 2

$$0.5\ N$$

$$1\ N$$

$$1.5\ N$$

$$2\ N$$

Solution

Question 3

$$\sqrt{3gr},2mg$$

$$\sqrt{2gr},mg$$

$$\sqrt{gr},mg$$

$$Zero$$ , $$Zero$$

Solution

Question 4

$$\dfrac{mg}{2}$$

$$mg$$

$$\dfrac{3mg}{2}$$

zero

Solution

Question 5

$$0$$

$$\dfrac{MgL}{2}$$

$$\dfrac{3MgL}{2}$$

$$\dfrac{5MgL}{2}$$

Solution

Question 6

2 : 3

3 : 2

1 : 1

1 : 2

Solution

Question 7

$$9\ N$$

$$7 \ N$$

$$5.5\ N$$

$$3.5 \ N$$

Solution

Question 8

$$\vec{\mathrm{P}_{1}}-\vec{\mathrm{P}}_{2}$$

$$\vec{\mathrm{P}}_{1}\sim\vec{\mathrm{P}}_{2}$$

$$\sqrt{\vec{\mathrm{P}_{1}}^{2}+\vec{\mathrm{P}}_{2}^{2}}$$

$$\sqrt{\vec{\mathrm{P}_{1}}^{2}-\vec{\mathrm{P}}_{2}^{2}}$$

Solution

Question 9

$$30^o$$

$$60^o$$

$$120^o$$

$$150^o$$

Solution

Question 10

is constant

increases linearly with time

decreases linearly with time

either increases or decreases linearly with time

Solution

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