Tag: work, energy and power

Questions Related to work, energy and power

A spring of natural length 3m and spring constant 9 N/m is having one end at origin and other end attached to a block of mass 1 kg. There is a wall at x=3m. At t=0 block is released from rest at x= 1 m. Collision of block with wall is elastic. Which of the following gives position of block with time :-

  1. $x=cos\left( 3t \right) $

  2. $x=3-2sin\left( 3t+\frac { \pi }{ 2 } \right) $

  3. $x=3-\left| 2cos\left( 3t \right) \right| $

  4. $x=3-2sin\left( 3t+\frac { 3\pi }{ 2 } \right) $


Correct Option: A

A steel ball moving with a velocity $\overline{v}$ collides with an identical ball originally at  rest. The velocity of the first ball after the collision is :

  1. $\left(-\dfrac{1}{2}\right)\overline{v}$

  2. $-\overline{v}$

  3. $\overline{v}$

  4. zero


Correct Option: D
Explanation:

Here, a steel ball moving with a velocity $\bar v$ collides with an identical ball originally at  rest. hence, masses of two steel balls are equal. For a head-on collision with a stationary object of equal mass, the projectile will come to rest and the target will move off with equal velocity, thus, the velocity of the first ball after the collision is zero.

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 : 

  1. 40 ms$^{-1}$

  2. 20 ms$^{-1}$

  3. 10 ms$^{-1}$

  4. 5 ms$^{-1}$


Correct Option: B
Explanation:

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 
$v _2 = 2u _1 = 2(10) = 20 ms^{-1}$

A block of mass 100$\mathrm { g }$ attached to a spring of stiffness 100$\mathrm { N } / \mathrm { m }$ is lying on a frictionless floor as shown. block is moved to compress the spring by 10 cm and released. If the collision with the wall is elastic then the time period of oscillations. (in seconds) 

  1. 0.133

  2. 13.3

  3. 0.26

  4. 0.3


Correct Option: A

Two particles of masses $ {m} _{1}, {m} _{2} $ movie with initial velocities $ u _{1} \text { and } u _{2} $.On collision, one of the particles get excited to higher level, after absorbing energy If final velocities of particles be $  v _{1}  $ and $  v _{2}  $ then we must have :

  1. $
    \dfrac{1}{2} m _{1} u _{1}^{2}+\dfrac{1}{2} m _{2} u _{2}^{2}=\dfrac{1}{2} m _{1} v _{1}^{2}+\dfrac{1}{2} m _{2} v _{2}^{2}-\varepsilon
    $

  2. $
    \dfrac{1}{2} m _{1} u _{1}^{2}+\dfrac{1}{2} m _{2} u _{2}^{2}+\varepsilon=\dfrac{1}{2} m _{1} v _{1}^{2}+\dfrac{1}{2} m _{2} v _{2}^{2}
    $

  3. $
    \dfrac{1}{2} m _{1}^{2} u _{1}^{2}+\dfrac{1}{2} m _{2}^{2} u _{2}^{2}-\varepsilon=\dfrac{1}{2} m _{1}^{2} v _{1}^{2}+\dfrac{1}{2} m _{2}^{2} v _{2}^{2}
    $

  4. $
    m _{1}^{2} u _{1}+m _{2}^{2} u _{2}-\varepsilon=m _{1}^{2} v _{1}+m _{2}^{2} v _{2}
    $


Correct Option: C
Explanation:

$\begin{array}{l} Total\, \, initial\, \, energy\, \, of\, \, two\, \, particles \ =\frac { 1 }{ 2 } { m _{ 1 } }{ u _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ u _{ 2 } }^{ 2 } \ Total\, \, final\, \, energy\, \, of\, \, two\, particles \ =\frac { 1 }{ 2 } { m _{ 1 } }{ v _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ v _{ 2 } }^{ 2 }+\in  \ U\sin  g\, \, energy\, \, conservation\, \, principle, \ \frac { 1 }{ 2 } { m _{ 1 } }{ u _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ u _{ 2 } }^{ 2 }=\frac { 1 }{ 2 } { m _{ 1 } }{ v _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ v _{ 2 } }^{ 2 }+\in  \ \therefore \frac { 1 }{ 2 } { m _{ 1 } }{ u _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ u _{ 2 } }^{ 2 }-\in =\frac { 1 }{ 2 } { m _{ 1 } }{ v _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ v _{ 2 } }^{ 2 } \end{array}$

Hence,
option $(C)$ is correct answer.

A moving sphere of mass m suffer a perfect elastic collision (not head on) with an  equally massive stationary sphere. after collision both fly off at angle $\theta $ value of which is :

  1. 0

  2. $\pi $

  3. indeterminate

  4. $\pi /2$


Correct Option: D

A rubber ball is bounced on the floor of a room which has its ceiling at a height of  $3.2{ m }$  from the floor. The ball hits the floor with a speed of  $10 m / { s },$  and rebounds vertically up. If all collisions simply reverse the velocity of the ball, without changing its speed, then how long does it take the ball for a round trip, from the moment it bounces from the floor to the moment it returns back to it ? Acceleration due to gravity is  $10 m / s ^ { 2 }.$

  1. $4 s$

  2. $2 s$

  3. $0.8 s$

  4. $1.2 s$


Correct Option: A

 A ball of mass 3 kg moving with a velocity of 4 m/s undergoes a perfectly- elastic collision with a stationary ball of mass m. After the impact is over, the kinetic energy of the 3 kg ball is 6 J. The possible value of m is/are :

  1. 1 kg only

  2. 1 kg , 9kg

  3. 1 kg, 6kg

  4. 6kg only


Correct Option: A

A proton of mass $m _p$ collides with a heavy particle. After collision proton bunches back with 4/9 of its intial kinetic energy. Collision is perfectly elastic. Find mass of heavy particle.

  1. 5 $m _p$

  2. 6 $m _p$

  3. 3 $m _p$

  4. 1.5 $m _p$


Correct Option: C

In an elastic collision the K.E of one body decreases by $100 J$. If the masses colliding bodies are in the ratio 3:4 the K.E of the other body increase by 

  1. $\dfrac{400}{3} J$

  2. $\dfrac{500}{3} J$

  3. $100 J$

  4. $0$


Correct Option: A