Work Energy and Power Test

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Work Energy and Power Test - 24

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Weekly Quiz Competition

Question 1
1 / -0

A ball with velocity $$9\ m/s$$ collides with another similar stationary ball. After the collision the two balls move in directions making an angle of $$30^o$$ with the initial direction. The ratio of the speeds of balls after the collision will be :

A
$$\displaystyle \dfrac{v_1}{v_2} = 1$$

B
$$\displaystyle \dfrac{v_1}{v_2} > 1$$

C
$$\displaystyle \dfrac{v_1}{v_2} < 1$$

D
$$\displaystyle \dfrac{v_1}{v_2} = 0$$

Solution

Let $$v_{1}$$ and $$ v_{2} $$ be the velocities of above and below particles respectively, as there is no external force .momentum is conserved.
conserving the momentum in vertical direction,

$$m v_{1}\sin(30) =mv_{2}\sin(30)$$

$$\Rightarrow v_{1}=v_{2}$$

$$\Rightarrow \dfrac{v_{1}}{v_{2}} =1$$
Question 2
1 / -0

A bullet weighing 10 g and moving at 300 ms$$^{-1}$$ strikes a 5 kg block of ice and drops dead. The ice block is sitting on frictionless level surface. The speed of the block, after the collision is :

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

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

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

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

Solution

Here, a bullet weighing 10 g and moving at 300 $$ms^{-1}$$ strikes a 5 kg block of ice and drops dead hence
$$m_1u_1 + m_2u_2 = m_2 v$$
$$10 \times 10^{-3} (300) + (5)(0) = 5 v$$
$$3000 \times 10^{-3} = 5 v$$
$$v = \dfrac{3}{5} = 0.6 ms^{-1}$$
Question 3
1 / -0

A toy car is tied to the end of an unstretched string of a length, $$a$$. When revolved, the toy car moves in a horizontal circle of radius $$2a$$ with time period, $$T$$. If it is now revolved in a horizontal circle of radius $$3a$$ with a period $$T'$$ with the same force, then

A
$$\displaystyle T' = \dfrac{\sqrt{3}}{2} T$$

B
$$\displaystyle T' = \sqrt {\dfrac{3}{2}} T$$

C
$$T' = T$$

D
$$\displaystyle T' = \dfrac{3}{2} T$$

Solution

$$\displaystyle m(2a)\left(\dfrac{2\pi}{T}\right)^2 = m(3a) \left(\dfrac{2\pi}{T'}\right)^2$$.

So, $$\dfrac{T'}{T}=\sqrt{\dfrac{3}{2}}$$
Question 4
1 / -0

In the elastic collision of heavy vehicle moving with a velocity 10 ms$$^{-1}$$ and a small stone at rest, the stone will fly away with a velocity equal to :

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

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

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

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

Solution

In the elastic collision between a heavy object and a very light object at rest, the velocity of particles after collision is
for heavy particle, $$v_1 = u_1$$
for light particle, $$v_2 = 2u_1 - u_2$$
since, $$u_2 = 0$$ hence,
$$v_2 = 2u_1$$
Therefore, the stone will fly away with a velocity equal to
$$v_2 = 2u_1 = 2(10) = 20 ms^{-1}$$
Question 5
1 / -0

A sphere of mass m, moving with a speed v, strikes a wall elastically at an angle of incidence $$\theta$$. If the speed of the sphere before and after collision is the same and the angle of incidence and velocity normally towards the wall the angle of rebound is equal to the angle of incidence and velocity normally towards the wall is taken as negative then, the change in the momentum parallel to wall is :

A
mv cos $$\theta$$

B
2mv cos $$\theta$$

C
-2mv cos $$\theta$$

D
zero

Solution

since the sphere collided elastically and there was no friction there was no impulse on the sphere along the wall.The only contact force acted was normal and that obviously was perpendicular to surface. NO change in momentum parallel to wall.
Question 6
1 / -0

A body of mass $$1\ kg$$ is rotating in a vertical circle of radius $$1\ m$$. What will be the difference in its kinetic energy at the top and bottom of the circle?
Take $$g = 10\ m/s^{2}$$

A
$$50\ J$$

B
$$30\ J$$

C
$$20\ J$$

D
$$10\ J$$

Solution

According to work energy theorem, $$\Delta K.E.=W$$ and here work is done by the gravitational force.
$$\Rightarrow \Delta K.E.=W = mg\ (2r) = 1 \times 10 \times 2(1)=20\ J$$
Question 7
1 / -0

A ball of mass $$m$$ moving with velocity $$v$$ collides elastically with wall and rebounds. The change in momentum of the ball will be :

A
$$4 mv$$

B
$$2 mv$$

C
$$mv$$

D
$$zero$$

Solution

Here, a ball of mass $$m$$ moving with velocity $$v$$ collides elastically with wall hence, momentum is
$$p_i = mv$$
The ball rebounds from wall hence, final momentum is
$$p_f = -mv$$
Change in momentum is
$$\Delta p = p_i - p_f$$
$$\Delta p = mv - (-mv) = 2mv$$
Question 8
1 / -0

What will be the potential energy of a body of mass 5 kg kept at a height of 10 m ?

A
50 J

B
0.5 J

C
500 J

D
25 J

Solution

Answer is C.

Potential energy is energy stored in an object. This energy has the potential to do work. Gravity gives potential energy to an object. This potential energy is a result of gravity pulling downwards. The gravitational constant, g, is the acceleration of an object due to gravity. This acceleration is about 10 meters per second on earth. The formula for potential energy due to gravity is PE = mgh. As the object gets closer to the ground, its potential energy decreases while its kinetic energy increases.
In this case, a body of mass 5 kg kept at a height of 10 m. So the potential energy is given as 5 * 10 *10 = 500 J.
Hence, the potential energy of a body of mass 5 kg kept at a height of 10 m is 500 J.
Question 9
1 / -0

The velocity of a particle at highest point of the vertical circle is $$\sqrt{3rg}$$. The tension at the lowest point, if mass of the particle is $$m$$, is

A
$$2\ mg$$

B
$$4\ mg$$

C
$$6\ mg$$

D
$$8\ mg$$

Solution

Let the velocity at lowest point be $$v$$. then according to law of conservation of energy.
energy at lowest point= energy at highest point

$$\Rightarrow \dfrac{1}{2}mv^2=mg(2r)+\dfrac{1}{2}m(\sqrt{3rg})^2 \\ \Rightarrow v=\sqrt{7rg}$$

At lowest point,

$$\dfrac { m{ v }^{ 2 } }{ r } =T-mg \\ \Rightarrow T=mg+ \dfrac { m{ v }^{ 2 } }{ r }=mg+ \dfrac { m (\sqrt{7rg} )^{ 2 } }{ r }$$

$$\Rightarrow T=mg+7mg=8mg $$
Question 10
1 / -0

Two objects that are moving along an xy-plane on a frictionless floor collide. Assume that they form a closed, isolated system. The following table gives some of the momentum components (in kilogram meters per second) before and after the collision. What are the mission values (a, b):
Before collision After collision
Object P$$_x$$ P$$_y$$ P$$_x$$ P$$_y$$
A -4 5 3 a
B b -2 4 2

A
10, 11

B
1, 11

C
5, 7

D
6, 4

Solution

Applying conservation of momentum in x-axis :
$$-4+ b = 3+4$$
$$\implies b = 11$$
Applying conservation of momentum in y-axis :
$$5-2 = a+2$$
$$\implies a = 1$$

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	Before collision		After collision
Object	P$$_x$$	P$$_y$$	P$$_x$$	P$$_y$$
A	-4	5	3	a
B	b	-2	4	2

Work Energy and Power Test - 24

Result

Work Energy and Power Test - 24

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SHARING IS CARING

Question 1

$$\displaystyle \dfrac{v_1}{v_2} = 1$$

$$\displaystyle \dfrac{v_1}{v_2} > 1$$

$$\displaystyle \dfrac{v_1}{v_2} < 1$$

$$\displaystyle \dfrac{v_1}{v_2} = 0$$

Solution

Question 2

60 ms$$^{-1}$$

3 ms$$^{-1}$$

6 ms$$^{-1}$$

0.6 ms$$^{-1}$$

Solution

Question 3

$$\displaystyle T' = \dfrac{\sqrt{3}}{2} T$$

$$\displaystyle T' = \sqrt {\dfrac{3}{2}} T$$

$$T' = T$$

$$\displaystyle T' = \dfrac{3}{2} T$$

Solution

Question 4

40 ms$$^{-1}$$

20 ms$$^{-1}$$

10 ms$$^{-1}$$

5 ms$$^{-1}$$

Solution

Question 5

mv cos $$\theta$$

2mv cos $$\theta$$

-2mv cos $$\theta$$

zero

Solution

Question 6

$$50\ J$$

$$30\ J$$

$$20\ J$$

$$10\ J$$

Solution

Question 7

$$4 mv$$

$$2 mv$$

$$mv$$

$$zero$$

Solution

Question 8

50 J

0.5 J

500 J

25 J

Solution

Question 9

$$2\ mg$$

$$4\ mg$$

$$6\ mg$$

$$8\ mg$$

Solution

Question 10

10, 11

1, 11

5, 7

6, 4

Solution

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