Tag: work, energy and power

Questions Related to work, energy and power

Two identical balls  $A$  and  $B$  having velocities of  $0.5\mathrm { m } / \mathrm { s }$  and  $- 0.3 \mathrm { m } / \mathrm { s }$  respectively collide elastically in one dimension. The velocities of  $B$  and  $\mathrm { A }$  after the collision respectively will be

  1. $0.3 \mathrm { m } / \mathrm { s } \text { and } 0.5 \mathrm { m } / \mathrm { s }$

  2. $- 0.5 \mathrm { m } / \mathrm { s } \text { and } 0.3 \mathrm { m } / \mathrm { s }$

  3. $0.5 \mathrm { m } / \mathrm { s } \text { and } - 0.3 \mathrm { m } / \mathrm { s }$

  4. $- 0.3 \mathrm { m } / \mathrm { s } \text { and } 0.5 \mathrm { m } / \mathrm { s }$


Correct Option: C

A particle of mass $ m _1 $ hits another particle of mass $ m _2 $ at rest with a velocity $ \overrightarrow { u }  $. The collision is head-on and elastic.If $ m _1 >> m _2 $, then after collision, the velocity of $ m _2 $ will be-

  1. $ \overrightarrow { u } $

  2. $ - \overrightarrow { u } $

  3. $ 2 \overrightarrow { u } $

  4. $ -2 \overrightarrow { u } $


Correct Option: C

Which of the following does no undergo elastic collision?

  1. When $ m _1 = m _2$ and $m _2 $ is stationary,there is maximum transfer of kinetic energy in head an collision

  2. When $ m _1 = m _2 $ is stationary,there is minimum transfer of momentum in head on collision

  3. When $ m _1 >> m _2 $ is stationary,after head on collision $ m _2 $ moves with twice the velocity of $ m _1 $

  4. When the collision is oblique and $ m _1 = m _2 with m _2 $ stationary,after the collision the particle move in opposite directions.


Correct Option: A

A perfectly elastic ball falls on a horizontal floor from a height in a time $t$. It will hit the floor again after a time $t'$. The ratio of $t'$ and t is 

  1. $1:1$

  2. $1:2$

  3. $2:1$

  4. $1:4$


Correct Option: B

IN a collision between two solid spheres, velocity of separation along the external forces act on the system of two sphers during impact 

  1. cannot be greater then velocity of approach

  2. cannot be less then velocity of approach

  3. cannot be equal to velocity of approach

  4. none of these


Correct Option: C

A sphere P of mass m and  velocity $\underset{V _{1}}{\rightarrow}$  undergoes an oblique and perfectly elastic collision with an identical sphere Q initially at rest.  The  angle $\Theta $  between the velocites of the spheres after the collision shall be

  1. 0

  2. $45^{\circ}$

  3. $90^{\circ}$

  4. $180^{\circ}$


Correct Option: C

A neutron collides head-on with a stationary hydrogen atom $( _1H^1)$ in ground state, then choose the correct statement (assume that mass of neutron and mass of $( _1H^1)$ atom is same)

  1. If kinetic energy of the neutron is less than $13.6eV$, collision must be elastic

  2. If kinetic energy of the neutron is less than $13.6eV$, collision must be inelastic

  3. Inelastic collision may take place only when initial kinetic energy of neutron is greater than $13.6eV$

  4. Perfectly inelastic collisin cannot take place.


Correct Option: A

Choose the correct statements from the following :

  1. the general form of Newton's second law of motion is $\vec{F} _{ext} = \vec m a$.

  2. a body can have energy and get no momentum.

  3. a body having momentum must necessarily have kinetic energy.

  4. the relative velocity of two bodies in a head-on elastic collision remains unchanged in magnitude and direction.


Correct Option: B,C
Explanation:

General form of 2nd law is $\vec { { F } _{ ext } } =m\vec { { a } _{ COM } } $
A body can have potential energy.
$KE=\dfrac { { p }^{ 2 } }{ 2m } $, so if a body has momentum it must have KE
its a fact that relative velocity in any type of collision changes in direction.

A point mass $M$ moving with a certain velocity collides with a stationary point mass $\dfrac{M}{2}$. The collision is elastic and one dimension. Let the ratio of the final velocities of $M$ and $\dfrac{M}{2}$ be $x$. The value of $x$ is :

  1. $2$

  2. $3$

  3. $\dfrac{1}{2}$

  4. $\dfrac{1}{4}$


Correct Option: D
Explanation:

$v _1=\dfrac{\left(M-\dfrac{M}{2}\right)}{\left(M+\dfrac{M}{2}\right)}u _1=\dfrac{u _1}{3}$


$\therefore v _2=\dfrac{4u _1}{3}$


$\therefore \dfrac{v _1}{v _2}=\dfrac{1}{4}$

A body of mass $M$ moving with a speed $u$ has a head-on collision with a body of mass $m$ originally at rest. If $M>>m$, the speed of the body of mass $m$ after collision will be nearly:

  1. $\dfrac{um}{M}$

  2. $\dfrac{uM}{m}$

  3. $\dfrac{u}{2}$

  4. $2u$


Correct Option: D