Tag: work, energy and power

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Questions Related to work, energy and power

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

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. 

  1. kinetic,kinetic

  2. elastic,kinetic

  3. elastic,elastic

  4. kinetic,elastic
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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?

  1. $\dfrac{1-e}{1+e}$
  2. $\dfrac{e-1}{e+1}$
  3. $\dfrac{1+e}{1-e}$
  4. $\dfrac{e+1}{e-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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,

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

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

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.

  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 $\omega$.
Reveal answer Fill a bubble to check yourself
A,C Correct answer
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.
Multiple choice physics work, energy and power collision of two rigid bodies energy and collisions understanding collisions

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.

  1. 8.4m/s

  2. 4.2m/s

  3. 2.1m/s

  4. 16.8m/s
Reveal answer Fill a bubble to check yourself
A Correct answer
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$

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

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 

  1. $t ^ { 1 / 2 }$
  2. $t ^ { 3 / 2 }$
  3. $t ^ { 5 / 2 }$
  4. $t ^ { 7 / 2 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.