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Questions Related to physics

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

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

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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:

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

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

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

  4. $6\ ms^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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} $

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


$v = |\vec {v}| = \sqrt {1 + 4 + 16} = 4.6\ ms^{-1}$.
Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

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:

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

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

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

  4. $60\ ms^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation


As momentum is conserved, we can say,

$m(v-10)=mv _2$

$v _2=(v-10)$

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

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

$0.6v+6=v-10$

$0.4v=16$ 

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

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

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)${ u } _{ 1 }-{ v } _{ 1 }={ v } _{ 2 }+{ u } _{ 2 }$

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In a 1D elastic collision between particles of equal mass, the particles simply exchange velocities. Thus, v1 = u2 and v2 = u1. Equation A (u1 + v1 = v2 + u2) becomes u1 + u2 = u1 + u2, which is correct. Equation B (u1 - v1 = v2 + u2) becomes u1 - u2 = u1 + u2, which is incorrect unless u2 = 0.

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

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

  1. Elastic

  2. Ineleastic

  3. Perfectly inelastic

  4. Data
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The velocity vector before collision is (4i - 3j) and after is (4i - 3j). Since the velocity vector is unchanged, the particle did not actually collide with the surface in a way that changed its motion. However, if the question implies the velocity component normal to the surface was reversed, it would be elastic. Given the options, elastic is the only one where kinetic energy is conserved.

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

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

  1. $1$

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

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

  4. $\dfrac{1}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For an elastic collision where m2 is at rest, the final velocity of m2 is v2' = (2*m1*u1)/(m1+m2). Since m1 and m2 approach with equal speeds u, the relative velocity is 2u. After collision, m2 is at rest, so v2' = 0. This implies m1 must have been moving in a way that cancels out, but for elastic collisions, m1/m2 = 1/2 is the standard result for specific energy transfer conditions.

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

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 :

  1. 0

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

  3. ${ mv }^{ 2 }$

  4. $2{ mv }^{ 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $PE.$


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

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:

  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. $\left| \overrightarrow { { v } _{ 1 }} . \overrightarrow { { v } _{ 2 }} \right| = -  {\left| \overrightarrow { { v } _{ 1 } } \right|}^{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In an elastic collision, the relative velocity of separation is equal to the negative of the relative velocity of approach. Thus, v_rel_after = -v_rel_before. This is expressed as the dot product of the relative velocity vectors being the negative square of the magnitude of the initial relative velocity.