Work Energy and Power Test

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Work Energy and Power Test - 18

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Question 1
1 / -0

A particle moves under the effect of a force $$F = C x$$ from $$x = 0$$ to $$ x = x_{1}$$. The work done in the process is (treat $$C$$ as a constant):

A
$$ \dfrac {C^{2}}{x^{2}_{1}}$$

B
$$Cx^{2}_{1}$$

C
$$ \dfrac {Cx^{2}_{1}}{2}$$

D
$$ \dfrac {C}{x^{2}_{1}}$$

Solution

Work Done $$\displaystyle =\int_{0}^{x_1} \vec{F}.d\vec{x}$$ $$\displaystyle=\int_{0}^{x_1}Cx.dx$$ $$= \dfrac {Cx^{2}_{1}}{2}$$
Question 2
1 / -0

A body of mass 5 kg moving with a speed of $$ 3 ms^{-1}$$ collides head on with a body of mass 3 kg moving in the opposite direction at a speed of $$2 ms^{-1}$$. The first body stops after the collision. Find the final velocity of the second body.

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

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

C
$$-9 ms^{-1}$$

D
$$30 ms^{-1}$$

Solution

$$\underline {Given:}$$
$$m_1=5kg$$
$$m_2=3kg$$
$$u_1=3ms^{-1}$$
$$u_2=-2ms^{-1}$$
Before collision, the 2nd body is moving backwards. Hence its velocity is negative.
$$v_1=0ms^{-1}$$
After collision, the 1st body stops moving. So its velocity is zero.

$$To$$ $$find:$$ $$v_2=$$ $$?$$

$$\underline {Solution:}$$

$$m_1 u_1+m_2 u_2$$ $$=$$ $$m_1 v_1+m_2 v_2$$

Substituting the values, we get,

$$(5\times3)$$ $$ +$$ $$(3\times(-2))$$ $$=$$ $$5\times 0$$ $$+$$ $$3\times v_2$$
$$\therefore15$$ $$-$$ $$6$$ $$=$$ $$3\times v_2$$
$$\therefore$$ $$9$$ $$=$$ $$3 v_2$$
$$\therefore$$ $$v_2$$ $$=$$ $$9$$ $$\div$$ $$3$$ $$=$$ $$3$$
$$\therefore$$ $$v_2$$ $$=$$ $$3ms^{-1}$$
Since we got a positive value for $$v_2,$$ it means that the 2nd body moves forwards after collision.
Velocity being a vector quantity, the direction is also important.

After collision, the 2nd body moves with a velocity $$3ms^{-1}$$ in the forward direction.
Question 3
1 / -0

A heavy steel ball of mass greater than 1 kg moving with a speed of 2m/ s collides head on with a stationary ping pong ball of mass less than 0.1 g. The collision is elastic. After the collision the ping pong ball moves approximately with a speed

A
$$ 2 m / s $$

B
$$4 m/ s$$

C
$$2\times10^{4}m / s$$

D
$$2\times10^{3}m / s$$

Solution

Since the body is much heavy these won't be much change in velocity
& $$e = 1$$

i.e., $$\dfrac{v-2}{0-2} = -1$$

$$\Rightarrow v = 4 m/s$$
Question 4
1 / -0

A ball of mass M moving with a velocity V collides head on elastically with another of same mass but moving with a velocity v in the opposite direction. After collision,

A
the velocities are exchanged between the two balls

B
both the balls come to rest

C
both of them move at right angles to the original line of motion

D
one ball comes to rest and another ball travels back with velocity 2V

Solution

Using conservation of linear momentum we have:
$$M_1u_1-M_2u_2=-M_1v_1+M_2v_2$$

$$MV-Mu_2=-Mv+Mv_2$$

$$V-u_2=-v+v_2$$.............(i)

and as for a perfect elastic collision we have $$e=1$$, thus we get

$$v_2+v=u_2+V$$.................(ii)

Using these two equations we get:
$$u_2=v$$ and $$v_2=V$$

Thus, the velocities are exchanged between the two balls.
Question 5
1 / -0

A heavier body moving with certain velocity collides head on elastically with a lighter body at rest. Then

A
smaller body continues to be in the same state of rest

B
smaller body starts to move in the same direction with same velocity as that of bigger body

C
the smaller body starts to move with twice the velocity of the bigger body in the same direction

D
the bigger body comes to rest

Solution

$$\textbf{Hint}$$: Apply Law of conservation of momentum
$$\textbf{Step 1:Assumptions}$$
Let us assume that there are two bodies. Let the first body has a mass of 'M' and initial velocity=$$V$$ and final velocity=$$V_1$$ (after collision).
Let the second body has a mass of 'm' and initial velocity=0 ; final velocity=$$V_2$$ (after collision)
$$\textbf{Step 2:Apply Law of conservation of Linear Momentum}$$
We know according to the law of conservation of momentum
Initial momentum=Final Momentum
$$MV+m(0)=MV_1+mV_2$$ $$(1)$$
$$\textbf{Step3:Coefficient of Restitution}$$
We know for elastic collision e=1
$$e=1=\dfrac{V_2-V_1}{V-0}$$
$$V=V_2-V_1$$
$$V_1=V_2-V$$ $$(2)$$
$$\textbf{Step 4:Solve above equations to obtain answer}$$
Substitute $$V_1=V_2-V$$ in $$(1)$$
we get
$$MV=M(V_1)+mV_2$$
$$MV=M(V_2-V)+mV_2$$
$$MV=MV_2-MV+mV_2$$
$$2MV=MV_2+mV_2$$
$$V_2=\dfrac{2VM}{M+m}$$
$$V_2=2V$$ if $$M>>m$$
Thus option C is correct
Question 6
1 / -0

A rain drop of mass (1/10) gram falls vertically at constant speed under the influence of the forces of gravity and viscous drag. In falling through 100 m, the work done by gravity is

A
0.98 J

B
0.098 J

C
9.8 J

D
98 J

Solution

work done = mgh
m = mass of drop = 0.1 g = 0.00001kg
g =9.8m/s
h = 100m
work done = $$0.00001 \times 9.8 \times 100$$
= 0.098 J
Question 7
1 / -0

A body of mass m is rotated at uniform speed along a vertical circle with the help of light string. If $$T_{1} and \ T_{2}$$ are tensions in the string when the body is crossing highest and lowest point of the vertical circle respectively, then which of the following expressions is correct?

A
$$T_{2}-T_{1}=6mg$$

B
$$T_{2}-T_{1}=4mg$$

C
$$T_{2}-T_{1}=2mg$$

D
$$T_{2}-T_{1}=mg$$

Solution

Centripetal Force is: $$\dfrac { m{ v }^{ 2 } }{ r } $$
At the highest point: $${ T }_{ 1 }=\dfrac { m{ v }^{ 2 } }{ r } -mg$$
At the lowest point: $${ T }_{ 2 }=\dfrac { m{ v }^{ 2 } }{ r } +mg$$
$$\therefore \quad { T }_{ 2 }-{ T }_{ 1 }=\dfrac { m{ v }^{ 2 } }{ r } +mg-(\dfrac { m{ v }^{ 2 } }{ r } -mg)=2mg$$
Question 8
1 / -0

A ball moving with a speed of 2.2 m/sec strikes an identical stationary ball. After collision the first ball moves at 1.1 m/sec at $$60^{0}$$ with the original line of motion. The magnitude and direction of the ball after collision is

A
$$5 m/sec,90^{0}$$

B
$$2 m/sec,60^{0}$$

C
$$\sqrt{39(1.1)}m/sec,30^{0}$$

D
$$10m/sec,60^{0}$$

Solution

Conserving momentum in X - direction

$$m \times 2.2 = 1.1 \times m \times cos 60 + v\:cos\:\theta \times m$$

$$2.2 = \dfrac{1.1}{2} + v\:cos\:\theta$$

$$ \dfrac{3.3}{2} = v\:cos\:\theta$$........................(1)

Similarly in Y - direction

$$v\:sin\:\theta = 1.1 sin \;60$$

$$v\:sin\:\theta = \dfrac{1.1\sqrt{3}}{2} $$........................(2)

from (1) & (2)

$$v = \sqrt{3.9(1.1)}m/s$$
$$ \theta = 30^{0}$$
Question 9
1 / -0

A 50 gm ball collides with another ball of mass 150gm, moving in its direction of motion, After collision the two balls move at a an angle $$30^{0}$$ with their initial direction. Ratio of their velocities after collision is

A
3 : 1

B
1 : 3

C
2 : 3

D
1 : 1

Solution

The momentum of the system is conserved in every direction. Since the initial vertical momentum is zero, the final vertical momentum is also zero.
Therefore,
$$ 50v_1 \times sin\,30^{\circ} = 150v_2 \times sin\,30^{\circ} $$

$$ \dfrac{v_1}{v_2} = \dfrac{3}{1}$$.
Question 10
1 / -0

A body is moving in a vertical circle such that the velocities of body at different points are critical. The ratio of velocities of body at angular displacements $$60^{0}$$ and $$120^{0}$$ from the lowest point is:

A
$$\sqrt{5}:\sqrt{2}$$

B
$$\sqrt{3}:\sqrt{2}$$

C
$$\sqrt{3}:1$$

D
$$\sqrt{2}:1$$

Solution

Given that velocities at different points are critical
so velocity at top point ($$T$$)
$$\Rightarrow$$ $${ V }_{ T }$$$$=\sqrt { gR } ...(1) $$
From the diagram
in right triangle $$\triangle OMA$$
$$\Rightarrow$$$$MA=OP=OP\cos { \theta } $$
At point $$A$$, $$\theta =60°$$
$${ \Rightarrow H }_{ A }=OG-OP\\ { \Rightarrow H }_{ A }=R-R\cos { 60° } =\frac { R }{ 2 } ...(2) $$
in right triangle $$\triangle OMB$$
$$\Rightarrow$$$$MB=QP=QP\cos { (180-\theta) } $$
At point $$B$$, $$\theta =120°$$
$${ \Rightarrow H }_{ B }=OG+OQ\\ { \Rightarrow H }_{ B }=R+R\cos { 60° } =\frac { 3R }{ 2 } ...(4) $$
Using energy conservation equation at point $$T$$ and $$A$$
$$\Rightarrow \frac { 1 }{ 2 } m{ { V }_{ T } }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ A } }^{ 2 }+mg{ H }_{ 1 }...(2)$$
putting the value of eqn (1), eqn (2) in eqn (4)
$$\Rightarrow \frac { 1 }{ 2 } m{ (\sqrt { gR } ) }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ A } }^{ 2 }+mg\frac { R }{ 2 } $$
$$\Rightarrow V_{ A }=2\sqrt { gR } $$
Using energy conservation equation at point $$T$$ and $$B$$
$$\Rightarrow \frac { 1 }{ 2 } m{ { V }_{ T } }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ B } }^{ 2 }+mg{ H }_{2}...(2)$$
putting the value of eqn (1), eqn (3) in eqn (4)
$$\Rightarrow \frac { 1 }{ 2 } m{ (\sqrt { gR } ) }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ B} }^{ 2 }+mg\frac { 3R }{ 2 } $$
$$\Rightarrow V_{ B }=\sqrt { 2gR } $$
Ratio of $${ V }_{ A }$$ and $${ V }_{ B }$$$$=\sqrt { 2 } :1$$
So the correct option is ($$D$$)

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Work Energy and Power Test - 18

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Work Energy and Power Test - 18

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SHARING IS CARING

Question 1

$$ \dfrac {C^{2}}{x^{2}_{1}}$$

$$Cx^{2}_{1}$$

$$ \dfrac {Cx^{2}_{1}}{2}$$

$$ \dfrac {C}{x^{2}_{1}}$$

Solution

Question 2

$$3 ms^{-1}$$

$$5 ms^{-1}$$

$$-9 ms^{-1}$$

$$30 ms^{-1}$$

Solution

Question 3

$$ 2 m / s $$

$$4 m/ s$$

$$2\times10^{4}m / s$$

$$2\times10^{3}m / s$$

Solution

Question 4

the velocities are exchanged between the two balls

both the balls come to rest

both of them move at right angles to the original line of motion

one ball comes to rest and another ball travels back with velocity 2V

Solution

Question 5

smaller body continues to be in the same state of rest

smaller body starts to move in the same direction with same velocity as that of bigger body

the smaller body starts to move with twice the velocity of the bigger body in the same direction

the bigger body comes to rest

Solution

Question 6

0.98 J

0.098 J

9.8 J

98 J

Solution

Question 7

$$T_{2}-T_{1}=6mg$$

$$T_{2}-T_{1}=4mg$$

$$T_{2}-T_{1}=2mg$$

$$T_{2}-T_{1}=mg$$

Solution

Question 8

$$5 m/sec,90^{0}$$

$$2 m/sec,60^{0}$$

$$\sqrt{39(1.1)}m/sec,30^{0}$$

$$10m/sec,60^{0}$$

Solution

Question 9

3 : 1

1 : 3

2 : 3

1 : 1

Solution

Question 10

$$\sqrt{5}:\sqrt{2}$$

$$\sqrt{3}:\sqrt{2}$$

$$\sqrt{3}:1$$

$$\sqrt{2}:1$$

Solution

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