Work Energy and Power Test

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Work Energy and Power Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

A particle of mass m has a velocity $$-v_{0}\hat{i}$$. while a second particle of same mass has a velocity $$v_{0}j$$. After the particles collide, first particle is found to have a velocity $$\dfrac{-1}{2}v_{0}\hat{i}$$ then the velocity of other particle is

A
$$\dfrac{-1}{2}v_{0}\hat{i}+v_{0}\hat{j}$$

B
$$\dfrac{1}{2}v_{0}\hat{i}+v_{0}\hat{j}$$

C
$$v_{0}\hat{i}+v_{0}\hat{j}$$

D
$$-v_{0}\hat{i}+v_{0}\hat{j}$$

Solution

Using momentum conservation
Final momentum $$=$$ Initial momentum

$$-\dfrac{m}{2}v_{0}\hat{i} + m(v_{x}\hat{i} + v_{y}\hat{j}) = -mv_{0}\hat{i} + mv_{0}\hat{j}$$

Comparing

$$v_{x} = -\dfrac{1}{2}v_{0}\hat{i}$$

$$v_{y} = v_{0}\hat{j}$$
Question 2
1 / -0

A wagon of mass 10 tons moving at a speed of 12 kmph collides with another wagon of mass 8 tons moving on the same track in the same direction at a speed of 10 kmph. If the speed of the first wagon decreases to 8 kmph. Find the speed of the other after collision

A
18 kmph

B
25 kmph

C
5 kmph

D
15 kmph

Solution

Conserving momemtum,
$$10 \times 12 + 8 \times10 = 10 \times 8 + 8 \times v$$
$$200 - 80 = 8 \times v$$
$$\Rightarrow v = 15 \ kmph$$
Question 3
1 / -0

A ball of mass 0.6kg attached to a light inextensible string rotates in a vertical circle of radius 0.75m such that it has speed of 5 m/s when the string is horizontal. Tension in string when it is horizontal on other side is:
$$(g=10ms^{-2})$$

A
$$30N$$

B
$$26N$$

C
$$20N$$

D
$$6N$$

Solution

Given: $$m=0.6 kg; R=0.75 m$$

Force equilibrium
$$T=\dfrac{mv^{2}}{R}$$

$$=\dfrac{(0.6)(5)^{2}}{(0.75)}$$

$$=20 N$$
Question 4
1 / -0

A simple pendulum is oscillating with an angular amplitude $$60^{0}$$ . If mass of bob is $$50\ g$$, the tension in the string at mean position is:
Consider: $$g=10ms^{-2}$$, length of the string, $$L = 1\ m$$.

A
$$0.5\ N$$

B
$$1\ N$$

C
$$1.5\ N$$

D
$$2\ N$$

Solution

Apply work energy theorem between A and B
$$mg L(1-cos\theta)=\dfrac{1}{2}mv^{2}$$

$$10\times 1\times (1-\dfrac{1}{2})=\dfrac{1}{2}mv^{2}$$

$$v^{2}=10$$

Then, the tension in, $$T=mg+\dfrac{mv^{2}}{R}$$

$$=\dfrac{50\times 10}{1000}+\dfrac{50}{1000}\times \dfrac{10}{1}$$

$$=1 N$$
Question 5
1 / -0

A 6 kg mass travelling at $$2.5$$ ms$$^{-1}$$ collides head on with a stationary 4 kg mass. After the collision the 6 kg mass travels in its original direction with a speed of $$1$$ ms$$^{-1}$$. The final velocity of 4 kg mass

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

B
$$2.25$$ ms$$^{-1}$$

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

D
$$0$$ ms$$^{-1}$$

Solution

Use momentum conservation,
$$m_1u_1+m_2u_2=m_1v_1+m_2v_2$$
But $$m_1=6$$kg$$, m_2=4$$kg, $$u_1=2.5$$ms$$^{-1}, u_2=0$$ms$$^{-1}, v_1=1$$ms$$^{-1}, v_2=?$$
$$ \Rightarrow 6 \times 2.5 +4\times (0) = 6 \times 1 + 4 \times v_2 $$
$$\Rightarrow 9 = 4v \Rightarrow v = 2.25 $$ ms$$^{-1}$$.
Question 6
1 / -0

A pilot of mass $$m$$ can bear a maximum apparent weight $$7$$ times of $$mg$$. The aeroplane is moving in a vertical circle. If the velocity of aeroplane is $$210\ m/s$$ while driving up from the lowest point of vertical circle, the minimum radius of vertical circle should be:

A
$$375\ m$$

B
$$420\ m$$

C
$$750\ m$$

D
$$840\ m$$

Solution

Given maximum apparent weight is $$7mg$$
Apparent weight at the lowest point is ($$mg + m \dfrac{v^2}{r} $$)
So, $$7mg = mg + m \dfrac{v^2}{r} $$
or, $$6mg = m \dfrac{v^2}{r} $$
or, $$r = \dfrac{v^2}{6g} $$
or, $$r = \dfrac{210^2}{6g} $$
or, $$ r = 750\ m$$
So , minimum radius of circle is $$750\ m $$
Question 7
1 / -0

A body of mass $$2\ kg$$ attached at one end of light string is rotated along a vertical circle of radius $$2\ m$$. If the string can withstand a maximum tension of $$140.6\ N$$, the maximum speed with which the stone can be rotated is:

A
$$22\ m/s$$

B
$$44\ m/s$$

C
$$33\ m/s$$

D
$$11\ m/s$$

Solution

$$ m= 2 Kg \quad R = 2m $$
$$T_{max} = 1406\ N $$
The tension is max at bottom point.
At bottom:
$$ T - mg = \dfrac{mv^{2}}{R}$$

$$ \dfrac{2(140.6- 2X9.81)}{2} = V^{2}$$

$${V = 10.99\ m/s}$$
Question 8
1 / -0

A body is revolving in a vertical circle of radius r such that the sum of its $$K.E.$$ and $$P.E.$$ is constant. If the speed of the body at the highest point is $$\sqrt{2gr}$$ then the speed of the body at the lowest point will be:

A
$$\sqrt{7gr}$$

B
$$\sqrt{6gr}$$

C
$$\sqrt{8gr}$$

D
$$\sqrt{9gr}$$

Solution

Applying WET between A and B.

$$\dfrac{m(\sqrt{2gR})^{2}}{2} + mg(2R) = \dfrac{m}{2}v_{0}^{2}$$

$$mgr + 2 mgR = \dfrac{m}{2}v_{0}^{2}$$

$$v_{0} = \sqrt{6gr}$$
Question 9
1 / -0

A simple pendulum is oscillating with an angular amplitude $$60^o$$. If $$m$$ is mass of bob and $$T_1$$, $$T_2$$ are tensions in the string, when the bob is at extreme position, mean position respectively then is:

A) $$T_1 = \dfrac{mg}{2}$$
B) $$T_2 = 2\ mg$$
C) $$T_1 = 0$$
D) $$T_2 = 3\ mg$$

A
A and B are true.

B
A and D are true.

C
B and C are true.

D
C and D are true.

Solution

Angular Magnitude is given $$\theta=60^o$$

As we see in figure, At extreme position:
$$T_A=mg\cos\theta$$
$$T_A=mg\cos 60^o$$

$$T_A=T_1=\dfrac{mg}{2}$$

As we see in figure, At middle position:

$$T_B=mg+F_c$$ where $$F_c$$ is centrifugal force

$$T_B=mg+\dfrac{mv^2}{R}$$ where $$v$$ is velocity of pendulum.

Using energy conservation for extreme position and middle position

$$v=\sqrt{gR}$$

putting values :

$$T_B=T_2=mg+\dfrac{mgR}{R}=2mg$$

Hence option A is correct.
Question 10
1 / -0

The bob of a simple pendulum at rest position is given a velocity $$V$$ in horizontal direction so that the bob describes vertical circle of radius equal to length of pendulum $$l$$ . If the tension in string is $$4$$ times weight of bob when the string is horizontal, the velocity of bob when it is crossing highest point of vertical circle is:

A
$$\sqrt{\dfrac{gl}{2}}$$

B
$$\sqrt{gl}$$

C
$$\sqrt{\dfrac{3gl}{2}}$$

D
$$\sqrt{2gl}$$

Solution

At pt B.

$$T = \dfrac{mv^{2}} {L} and T = 4mg.$$

$$4\dfrac{mg}{m}L = v^{2}$$

$$ V= \sqrt{4gL}$$

Now WET between B and C.

$$mgL = \dfrac{1}{2}mv^{2} - \dfrac{m}{2}v_{0}^{2}$$

$$mgL = 2mgL - \dfrac{m}{2}v_{0}^{2}$$

$$v_{0}= \sqrt{2gL}$$

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Work Energy and Power Test - 19

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Work Energy and Power Test - 19

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Question 1

$$\dfrac{-1}{2}v_{0}\hat{i}+v_{0}\hat{j}$$

$$\dfrac{1}{2}v_{0}\hat{i}+v_{0}\hat{j}$$

$$v_{0}\hat{i}+v_{0}\hat{j}$$

$$-v_{0}\hat{i}+v_{0}\hat{j}$$

Solution

Question 2

18 kmph

25 kmph

5 kmph

15 kmph

Solution

Question 3

$$30N$$

$$26N$$

$$20N$$

$$6N$$

Solution

Question 4

$$0.5\ N$$

$$1\ N$$

$$1.5\ N$$

$$2\ N$$

Solution

Question 5

$$1$$ ms$$^{-1}$$

$$2.25$$ ms$$^{-1}$$

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

$$0$$ ms$$^{-1}$$

Solution

Question 6

$$375\ m$$

$$420\ m$$

$$750\ m$$

$$840\ m$$

Solution

Question 7

$$22\ m/s$$

$$44\ m/s$$

$$33\ m/s$$

$$11\ m/s$$

Solution

Question 8

$$\sqrt{7gr}$$

$$\sqrt{6gr}$$

$$\sqrt{8gr}$$

$$\sqrt{9gr}$$

Solution

Question 9

A and B are true.

A and D are true.

B and C are true.

C and D are true.

Solution

Question 10

$$\sqrt{\dfrac{gl}{2}}$$

$$\sqrt{gl}$$

$$\sqrt{\dfrac{3gl}{2}}$$

$$\sqrt{2gl}$$

Solution

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