Tag: mechanics

Quizzes (24) Questions (118)

Questions Related to mechanics

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

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.
Reveal answer Fill a bubble to check yourself
B,C Correct answer
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.

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

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

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


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


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

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

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. $2u$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

When a very massive body (M) hits a stationary light body (m) head-on, the light body is projected forward with twice the velocity of the massive body.

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

A ball moving with a certain velocity hits another identical ball at rest. If the plane is frictionless and collision is elastic, the angle between the directions in which the balls move after collision, will be

  1. $30^{o}$
  2. $60^{o}$
  3. $90^{o}$
  4. $120^{o}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For an elastic collision between two identical masses where one is initially at rest, the angle between the final velocities is 90 degrees.

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

Two perfectly elastic objects $A$ and $B$ of identical mass are moving with velocities $15\ m/s$ and $10\ m/s$ respectively collide along the direction of line joining them. Their velocities after collision are respectively:

  1. $10\ m/s, 15\ m/s$
  2. $20\ m/s, 5\ m/s$
  3. $0\ m/s, 25\ m/s$
  4. $5\ m/s, 20\ m/s$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$15m+10m=mv _1+mv _2$

$25=v _1+v _2$...............(i)
and $\dfrac{v _2-v _1}{u _1-u _2}=1$
$\Rightarrow \dfrac{v _2-v _1}{15-10}=1$
$\Rightarrow v _2-v _1=5$............(ii)
$v _1+v _2=25$
$\dfrac{v _2-v _1=5}{2v _2=30}$
$\therefore v _2=15m/s,v _1=10m/s$

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

A body of mass $8\ kg$ collides elastically with a stationary mass of $2\ kg$. If initial $KE$ of moving mass be $E$, the kinetic energy left with it after the collision will be:

  1. $0.80\ E$
  2. $0.64\ E$`
  3. $0.36\ E$
  4. $0.08\ E$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The fraction of kinetic energy retained by the incident mass m1 after an elastic collision with stationary mass m2 is ((m1 - m2)/(m1 + m2))^2. Here, m1=8, m2=2, so the fraction is ((8-2)/(8+2))^2 = (6/10)^2 = 0.36.

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

If two bodies $A$ and $B$ of definite shape (dimensions of bodies are not ignored) $A$ is moving with speed of $10\ m/s$ and $B$ is in rest. They collide elastically. Then;

  1. body $A$ comes to rest and $B$ moves with speed of $10\ ms$

  2. they may move perpendicular to each other
  3. $A$ and $B$ may come to rest

  4. they must move perpendicular to each other
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In a general elastic collision between two bodies, the conservation of momentum and energy allows for various scattering angles depending on the impact parameter. They do not necessarily move perpendicular to each other unless the masses are equal and one is at rest.

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

A ball of mass $m$ moving with velocity $v$ collides elastically with another ball of identical mass coming from opposite direction with velocity $2v$. Their velocities after collision will be :

  1. $-v,2v$
  2. $-2v,v$
  3. $v,-2v$
  4. $2v,-v$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$mv-2mv=mv _{1}+mv _{2}$
$-v=v _{1}+v _{2}$...........(1)

and $\dfrac{v _{2}-v _{1}}{v+2v}=1\Rightarrow v _{2}-v _{1}=3v$............(2)
Solving eqs. $(i)$ and $(ii)$,
$v _{2}-v _{1}=3v$
$\dfrac{v _{2}+v _{1}=-v}{2v _{2}=2v}$
$\therefore v _{2}=v$ and $-v _{1}=2v\therefore v _{1}=-2v$

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

Two solid rubber balls $A$ and $B$ having masses $200\ g$ and $400\ g$ respectively are moving in the opposite direction. A velocity of $A$ which is equal to $0.3\ m/s$. After the collision the two balls come to rest when the velocity of $B$ is 

  1. $0.15\ m/s$
  2. $1.5\ m/s$
  3. $-0.15\ m/s$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

From conservation of linear momentum:

               $P _i=P _f$                  $(P _f=0)$

$m _{A}v _{A}+m _{B}v _{B}=0$
$v _{B}=-\dfrac{m _{A}v _{A}}{m _{B}}=-\dfrac{200\times 10^{-3}\times 0.3}{400\times 10^{-3}}$
$V _B=-\dfrac{60}{400}=-0.15$