Tag: work, energy and power

Questions Related to work, energy and power

If a ball is dropped from rest, it bounces from the floor. The coefficient of restitution is $0.5$ and the speed just before the first bounce is $5\ m/sec$. The total time taken by the ball to come to rest is:

physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

  1. $2\ sec$

  2. $1\ sec$

  3. $0.5\ sec$

  4. $0.25\ sec$

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Correct Option: C

An athelete diving off a high spring board can perform a variety of physical moments in the air before entering the water below. Which one of the following parameters will remain constant during the fall? The athelete's:

physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

  1. linear velocity

  2. linear momentum

  3. moment of inertia

  4. angular velocity

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Correct Option: D

In an elastic collision, kinetic energy of the relative motion is converted into the ____ energies of two momentarily compressed bodies, and then is converted back into the _____ energy. Fill in the blanks. 

physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

  1. kinetic,kinetic

  2. elastic,kinetic

  3. elastic,elastic

  4. kinetic,elastic

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Correct Option: B
Explanation:

During elastic collision, at first both the bodies get deformed, so total kinetic energy is converted into theie elastic energies and then they regain their original shape (when moving apart with different velocities) suggesting that the elastic energy is converted back into the kinetic energy.

State whether the given statement is True or False :

In an elastic collision, the net kinetic energy of the two colliding bodies is conserved.

physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

  1. True

  2. False

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Correct Option: A
Explanation:

In an elastic collision, the net kinetic energy as well as linear momentum of the two colliding bodies is conserved. The given statement is true.

A ball of mass m moving with a constant velocity u strikes against a ball of same mass at rest. If e is the coefficient of restitution, then what will be the ratio of velocity of two balls after collision?

physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

  1. $\dfrac{1-e}{1+e}$

  2. $\dfrac{e-1}{e+1}$

  3. $\dfrac{1+e}{1-e}$

  4. $\dfrac{e+1}{e-1}$

Show answer
Correct Option: A
Explanation:

A. $\dfrac{1-e}{1+e}$


Given,


$m _1=m _2=m$ (say)

$u _1=u $, $u _2=0$

let, $v _1=$ velocity of ball 1 after collision

      $v _2=$ velocity of ball 2 after collision

The coefficient of restitution,

$e=\dfrac{v _2-v _1}{u _1-u _2}$

$eu=v _2-v _1$. . . . . . .(1)

By the conservation of Linear momentum,

$m _1u _1+m _2u _2=m _1v _1+m _2v _2$ 

$u=v _1+v _2$. . . . . . . . .(1)

By solving equation (1) and (2), we get

$v _1=\dfrac{(1-e)u}{2}$

$v _2=\dfrac{(1+e)u}{2}$

The ratio of velocity of two ball,

$\dfrac{v _1}{v _2}=\dfrac{1-e}{1+e}$

A sphere $A$ moving with speed $u$ and rotating with an angular velocity $\omega$ makes a head-on elastic collision with an identical stationary sphere $B$. There is no friction between the surfaces of $A$ and $B$. Choose the correct alternative(s). Discard gravity.

physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

  1. $A$ will stop moving but continue to rotate with an angular velocity $\omega$

  2. $A$ will come to rest and stop rotating

  3. $B$ will move with speed $u$ without rotating

  4. $B$ will move with speed $u$ and rotate with an angular velocity $\omega$.

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Correct Option: A,C
Explanation:
Let $m$ be the mass of sphere and $v _1$ and $v _2$ be the velocities of the spheres A and B respectively after the collision.
Applying conservation of linear momentum:        $P _i = P _f$
$m  u + 0 = mv _1  + mv _2           \implies v _1 + v _2 = u$      ........(1)
Also $\dfrac{v _2- v _1}{ u-0} = -1         \implies v _1 - v _2 = -u$  
On solving we get,       $v _1 = 0 $   and $v _2 = u$
Thus A will stop and B will move with $u$
Also as there is no friction between the A and B, thus there will be no torque. Hence angular velocities of the respective spheres must remains the same as it was initially.
So, A will continue to rotate with $w$ whereas B will not rotate.

A 20 g ball is fired horizontally toward a 100 g ball that is hanging motionless from a 1.0-m-long string. The balls undergo a head-on, elastic collision, after which the 100 g ball swings out to a maximum angle of 50 degrees. Determine the initial speed of the 20 g ball.

physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

  1. 8.4m/s

  2. 4.2m/s

  3. 2.1m/s

  4. 16.8m/s

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Correct Option: A
Explanation:

At first we can find the final velocity of the $100\ g$ ball.

$ \dfrac { { m } _{ 2 }{ { v } _{ 2 } }^{ 2 } }{ 2 } =mgh={ m } _{ 2 }gl(1-cos\alpha )$

$ \implies { v } _{ 2 }=\sqrt { 2gl\left( 1-cos\alpha  \right)  } =\sqrt { 2*9.81*1.1(1-cos{ 50 }^{ o }) } =2.78m/s $

Now applying conservation of the momentum principle, we get:

$ { m } _{ 1 }{ V } _{ 1 }+{ m } _{ 2 }{ V } _{ 2 }={ m } _{ 1 }{ v } _{ 1 }+{ m } _{ 2 }{ v } _{ 2 }$

$ { v } _{ 1 }=\dfrac { { m } _{ 1 }{ V } _{ 1 }-{ m } _{ 2 }{ v } _{ 2 } }{ { m } _{ 1 } } ={ V } _{ 1 }-{ v } _{ 2 }\dfrac { { m } _{ 2 } }{ { m } _{ 1 } } $

Principle of kinetic energy conservation , we get 

$ \dfrac { 1 }{ 2 } { m } _{ 1 }{ { { V } _{ 1 } }^{ 2 } }+\dfrac { 1 }{ 2 } { m } _{ 2 }{ { V } _{ 2 } }^{ 2 }=\dfrac { 1 }{ 2 } { m } _{ 1 }{ { v } _{ 1 } }^{ 2 }+\dfrac { 1 }{ 2 } { m } _{ 2 }{ { v } _{ 2 } }^{ 2 }$


$ { v } _{ 1 }={ V } _{ 1 }-{ v } _{ 2 }\dfrac { { m } _{ 2 } }{ { m } _{ 1 } } \ { V } _{ 2 }=0$

$\implies { V } _{ 1 }=\dfrac { { { { { m } _{ 1 } }^{ 2 }v } _{ 1 } }^{ 2 }+{ m } _{ 2 }{ m } _{ 1 }{ { v } _{ 2 } }^{ 2 } }{ 2{ m } _{ 1 }{ v } _{ 2 }{ m } _{ 2 } } =\dfrac { v _{ 2 }\left( { m } _{ 2 }+{ m } _{ 1 } \right)  }{ 2{ m } _{ 1 } } $

$ \implies  \dfrac { 2.78\left( 0.1+0.024 \right)  }{ 2*0.024 } =8.4m/s$

An energy of 4 kJ causes a displacement of 64 m in 2.5 s. The power delivered is

power work and power work, energy and power physics energy and its forms

  1. 16 W

  2. 160 W

  3. 1600 W

  4. 16000 W

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Correct Option: C
Explanation:

Given that


Energy, $E=4\ kJ=4000\ J$
Time taken , $t=2.5\ secs$

Power delivered , $P=\dfrac Et=\dfrac{4000}{2.5}= 1600\ W$

A constant power is delivered to a body moving along a straight line. the distance travelled by the body in time t is proportional to 

power work and power work, energy and power physics energy and its forms

  1. $t ^ { 1 / 2 }$

  2. $t ^ { 3 / 2 }$

  3. $t ^ { 5 / 2 }$

  4. $t ^ { 7 / 2 }$

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Correct Option: B
Explanation:

$\begin{array}{l} power\, P\, =Fv=mav \ acceleration\, a=\frac { S }{ { { t^{ 2 } } } } \, and\, velocity\, v=\frac { S }{ t } , \ where\, \, S\, \, is\, the\, dis\tan  ce\, moved \ Since\, \, the\, power\, P\, and\, mass\, m\, are\, cons\tan  t,\, we\, \, get \ P=\frac { { mS } }{ { { t^{ 2 } } } } \times \frac { S }{ t } \to { S^{ 2 } }={ t^{ 2 } }\to S\propto { t^{ \frac { 3 }{ 2 }  } } \end{array}$

Hence,
option $B$ is correct answer.

In the above question, if the work done on the system along the curved path $ba$ is $52\ J$, heat absorbed is 

power work and power work, energy and power physics energy and its forms

  1. $-140\ J$

  2. $-172\ J$

  3. $140\ J$

  4. $172\ J$

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Correct Option: A