Tag: collisions in one dimension

Questions Related to collisions in one dimension

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

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. $120^{o}$


Correct Option: C

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. $5\ m/s, 20\ m/s$


Correct Option: A
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)
$v _1+v _2=25$
$\dfrac{v _2-v _1=5}{2v _2=30}$
$\therefore v _2=15m/s,v _1=10m/s$

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. $0.08\ E$


Correct Option: C

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


Correct Option: B

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. $2v,-v$


Correct Option: B
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 $-v _{1}=2v\therefore v _{1}=-2v$