Tag: physics

Questions Related to physics

An astronomical refractive telescope has an objective of focal length 20 m and an eyepiece of focal length 2 cm. then

physics observing space: telescopes space research and satellites space travel space exploration and forms of light

  1. the magnification is 1000

  2. the length of the telescope tube is 20.02 m

  3. the image formed is inverted

  4. all of these

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Correct Option: D
Explanation:

Here, $f _0 = 20 \,m \,\, and \,\, f _e = 2 \, cm = 0.02 \, m$ In normal adjustment, Length of telescope tube, $L = f _0 + f _e = 20 +0.02 = 20.02m$
and magnification, $m = \dfrac{f _0}{f _e} = \dfrac{20}{0.02} = 1000$
The image formed is inverted with respect to the object.

A particle of mass $1\ g$ moving with a velocity $\vec {v _{1}} = 3\hat {i} - 2\hat {j} ms^{-1}$ experiences a perfectly in elastic collision with another particle of mass $2\ g$ and velocity $\vec {v _{2}} = 4\hat {j} - 6\hat {k} ms^{-1}$. The velocity of the particle is:

collisions in one dimension collisions work, energy and power mechanics physics

  1. $2.3\ ms^{-1}$

  2. $4.6\ ms^{-1}$

  3. $9.2\ ms^{-1}$

  4. $6\ ms^{-1}$

Show answer
Correct Option: B
Explanation:

From conservation of momentum
$m _{1}\vec {v _{1}} + m _{2}\vec {v _{2}} = (m _{1} + m _{2})\vec {v}$


$1\times (3\hat {i} - 2\hat {j}) + 2\times (4\hat {j} - 6\hat {k}) = (1 + 2)\vec {v}$

$\Rightarrow 3\hat {i} + 6\hat {j} - 12\hat {k} = 3\vec {v} $

$\Rightarrow \vec {v} = \hat {i} + 2\hat {j} - 4\hat {k}$


$v = |\vec {v}| = \sqrt {1 + 4 + 16} = 4.6\ ms^{-1}$.

A ball P moving with a speed of $v \ ms^{-1}$ collides directly with another identical ball Q moving with a speed $10\ ms^{-1}$ in the opposite direction. P comes to rest after the collision. If the coefficient of restitution is 0.6, the value of $v$ is:

collisions in one dimension collisions work, energy and power mechanics physics

  1. $30\ ms^{-1}$

  2. $40\ ms^{-1}$

  3. $50\ ms^{-1}$

  4. $60\ ms^{-1}$

Show answer
Correct Option: B
Explanation:


As momentum is conserved, we can say,

$m(v-10)=mv _2$

$v _2=(v-10)$

$e=\dfrac{v _2-v _1}{u _1+u _2}=\dfrac{(v-10)-0}{(v+10)}$

$0.6=\dfrac{v-10}{v+10}$

$0.6v+6=v-10$

$0.4v=16$ 

$v=40\ ms^{-1}$

A ball is dropped from a $45\ m$ high tower while another is simultaneously thrown upward from the foot at $20\ m/s$, along the same vertical line. If the collision is perfectly elastic, first ball reaches ground after time-

collisions in one dimension collisions work, energy and power mechanics physics

  1. $2s$

  2. $3s$

  3. $4s$

  4. $5s$

Show answer
Correct Option: A

A body of mass $4m$ at rest explodes into three pieces. Two of the pieces each of mass $m$ move with a speed $v$ each in mutually perpendicular directions. The total kinetic energy released is:

collisions in one dimension collisions work, energy and power mechanics physics

  1. $\cfrac{1}{2}m{v}^{2}$

  2. $m{v}^{2}$

  3. $\cfrac{3}{2}m{v}^{2}$

  4. $\cfrac{5}{2}m{v}^{2}$

Show answer
Correct Option: A

A particle of mass m moving with velocity ${u} _{1}$ collides elastically with particle of same mass moving with velocity ${u} _{2}$ in the same direction. After collision their speeds are ${v} _{1}$ and ${v} _{2}$ respectively then-
(A) ${ u } _{ 1 }+{ v } _{ 1 }={ v } _{ 2 }+{ u } _{ 2 }$
(B)${ u } _{ 1 }-{ v } _{ 1 }={ v } _{ 2 }+{ u } _{ 2 }$

collisions in one dimension collisions work, energy and power mechanics physics

  1. Both the equations A and B are correct

  2. Both the equations A and B are incorrect

  3. Equation A is correct but not B

  4. Equation B is correct but not A

Show answer
Correct Option: C

A particle of mass $1\ kg$ moving with a velocity of $(4\hat {i}-3\hat {j})m/s$ collides with a fixed surface. After the collision velocity of the particle is $(4\hat {i}-3\hat {j})m/s$. Collision is

collisions in one dimension collisions work, energy and power mechanics physics

  1. Elastic

  2. Ineleastic

  3. Perfectly inelastic

  4. Data

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Correct Option: A

Two masses $m _{1}$ and $m _{2}$, approaches each other with equal speeds and collide elastically. After collision $m _{2}$ comes to rest. Then $m _{1}$/$m _{2}$ is

collisions in one dimension collisions work, energy and power mechanics physics

  1. $1$

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

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

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

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Correct Option: B

Two identical balls each of mass in are moving in opposite direction with a speed v. if they collide elastically maximum potentail energy stored in the ball is :

collisions in one dimension collisions work, energy and power mechanics physics

  1. 0

  2. $\dfrac { 1 }{ 2 } { mv }^{ 2 }$

  3. ${ mv }^{ 2 }$

  4. $2{ mv }^{ 2 }$

Show answer
Correct Option: A
Explanation:

Net momentum before collision will be $mv+(-mv)=0$, $negative$ because from $opposite $ direction.

so after the collision they will get stopped to make the net momentum again $zero$ and whole energy will be get stored
 as $PE$. ($inelastic $ $ collision$)

There is one other possibility too that is they $exchange$ their velocities so that again the net momentum will become
 $zero.$ $elastic$ $ collision$ .
In elastic collision there is no loss in $KE$  so $no$ storage of $PE.$


Two particles moving initially in the same direction undergo a one dimensional,elastic collision. Their relative velocities before and after the collision are $\overrightarrow { { v } _{ 1 } } $ and $\overrightarrow { { v } _{ 2 } } $. Then:

collisions in one dimension collisions work, energy and power mechanics physics

  1. $\left| \overrightarrow { { v } _{ 1 } } \right| = \left| \overrightarrow { { v } _{ 2 } } \right| $

  2. $\overrightarrow { { v } _{ 1 } } = - \overrightarrow { { v } _{ 2 } }$ only if the two are of equal mass.

  3. $\overrightarrow { { v } _{ 1 } } = -\overrightarrow { { v } _{ 2 } } = {\left| \overrightarrow { { v } _{ 1 } } \right|}^{2}$

  4. $\left| \overrightarrow { { v } _{ 1 }} . \overrightarrow { { v } _{ 2 }} \right| = -  {\left| \overrightarrow { { v } _{ 1 } } \right|}^{2}$

Show answer
Correct Option: D