Collisions in one dimension - class-XI

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

As momentum is conserved, we can say,

$m(v-10)=mv _2$

$v _2=(v-10)$

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

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

$0.6v+6=v-10$

$0.4v=16$ 

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

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.

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.

For an elastic collision where m2 is at rest, the final velocity of m2 is v2' = (2m1u1)/(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.

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

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.

The Coefficent of Restitution is a measure of the "bounciness" of a collision between two objects: how much of the kinetic energy remains for the objects to rebound from one another vs. how much is lost as heat, or work done deforming the objects.

The coefficient, e is defined as the ratio of relative speeds after and before an impact, taken along the line of the impact:
$e=\dfrac { Speed\quad of\quad separation }{ Speed\quad of\quad approach } $

(i)   For perfectly elastic collision $e = 1$
(ii)  For perfectly inelastic collision $e = 0$
(iii) For other collision $0 \lt e \lt 1$

Relative speed of the ball $\displaystyle =\left( \upsilon +u \right) $
Speed after rebouncing $\displaystyle =-\left( \upsilon +u \right) $
Now, $\displaystyle F=m\frac { \Delta \upsilon  }{ \Delta t } =\frac { m }{ t } \left[ \left( \upsilon +u \right)  \right] -[-\left[ \left( \upsilon +u \right)  \right]] $
$\displaystyle =\frac { 2m }{ t } \left( \upsilon +u \right) $

The speed of the ball remains the same $v _0$ before and after the collision with wall.

Since the collision is inelastic, applying momentum conservation for inelastic collisions,
$ mv + m(-v) = (m+m)V $
$ V = 0 $.

The coefficient of restitution is defined as the ratio of the relative velocity of separation to that of approach, in a situation of two objects colliding with each other. 

In a collision system where $m _1$ moves with $u _1$ initially and $m _2$ is at rest with $m _2 >>> m _1$.
Let the final velocity of $m _1$ be $v _1$ and $m _2$ be $v _2$.
Assumption: Let the collision be elastic. Using linear momentum conservation and equation for coefficient of restitution,
$ m _1u _1 = m _2v _2 + m _1v _1 $
$ v _2 - v _1 = u _1 $

$mv+m(-v)=m{ v } _{ 1 }+m{ v } _{ 2 }=0\Rightarrow { v } _{ 1 }+{ v } _{ 2 }=0\$
$ \dfrac { { v } _{ 1 }-{ v } _{ 2 } }{ v-(-v) } =1\Rightarrow { v } _{ 1 }-{ v } _{ 2 }=2v\ $

$e=\dfrac{v-1}{2.5}$

$=\dfrac{1.25}{2.5}=\dfrac{1}{2}$

$\cfrac{1}{2}mv _o^2=\cfrac{1}{2}Kx^2+\cfrac{1}{2}\times m(0.25)v _o^2\quad equation (1)$

Let speed of balls are $v _1$ and $v _2$.

External force = $F _{ext}=\dfrac{\Delta p}{\Delta t}=((m _{1}v _{1}^{'}+m _{2}v _{2}^{'})-(m _{1}v _{1}+m _{2}v _{2}))/2t _{0}-0$

To excite the atom to the n=3 level (energy -1.51 eV from -13.6 eV), the energy required is 12.09 eV. If it emits a 2.55 eV photon (n=3 to n=2), the remaining energy must be 10.2 eV (n=2 to n=1). The minimum energy K must account for the excitation and the recoil energy of the atom.

In an elastic collision with a rigid wall, the particle reverses its velocity. Since the spring potential energy is conserved and the collision is elastic, the total energy remains constant. The particle will return to the equilibrium position with the same speed u0 it had initially.

Since the velocity before and after the collision change hence momentum of the ball will change. So (a) is not true.  

For identical spheres colliding elastically, they exchange velocities. If they cross without touching, they continue with their original velocities. If they collide, they also end up with the same final velocities as if they had passed through each other. Thus, the final state is indistinguishable.

u is initial velocity 

In a perfectly inelastic collision between identical masses, the spheres stick together and move with a common velocity after collision. Since they combine and move as one object, they travel in the same direction, making the angle between their motion directions 0°.

For an elastic collision between identical masses where one is at rest, the particles move at 90 degrees to each other after the collision, provided the collision is not head-on.

In a head-on elastic collision between two bodies of equal mass, they exchange velocities. If the second body was at rest, the first body comes to rest and the second moves with the initial velocity of the first. The angle between them is 180 degrees if we consider the path of the first body after it stops (which is not applicable) or 0 degrees. However, in general 2D elastic collisions of equal masses, they move at 90 degrees. The option 180 degrees is likely a distractor or refers to a specific 1D case.

In a head-on elastic collision between two bodies of equal mass, the velocities simply swap. Before collision, ball 1 has velocity V and ball 2 has velocity -2V (moving towards it). After the elastic collision, their velocities become -2V and V respectively.

Solving the equation of conservation of momentum give us that the relative velocity of approach is equal to the relative velocity of separation. Hence coefficient of restitution is 1.
Which means collision is elastic. 

Using conservation of momentum in two dimensions for equal masses, the sum of x and y velocity components before collision must equal the sum after collision. Initial total V_x = 50 - 40 = 10, and since the first ball comes to rest (V_x = 0, V_y = 0), the second ball must carry the total momentum, giving V_x = 10 and V_y = 30 - 0 = 30 cm/s.

When m1 = m2 and m2 is stationary, the first particle transfers all its kinetic energy to the second particle in a head-on collision. This is the maximum possible transfer.

When identical balls collide $elastically $ they just $exchange$ their $SPEEDS$ and get reversed.

The impact parameter is the perpendicular distance between the path of the projectile and the center of the target body. For a direct head-on collision, the centers of the two colliding bodies are directly aligned along the line of motion, making the impact parameter zero.

If $ { v } _{ 1 }$ is relative velocity before collision.
if ${ v } _{ 2 } $ is relative velocity before collision.
$e\le 1\$
$ e=\dfrac { { v } _{ 1 } }{ { -v } _{ 2 } } $

Let  $u=$ speed of A before impact. Thus,  $p=mu$.
Let $v _1, v _2 = $ speeds of  $A$ and $B$ after impact.
$u = v _1 + v _2 $ and $v _1 - v _2 = - eu$
$u = v _1 + v _2$ and $v _1 - v _2 = - eu$

For a head-on collision with a stationary object of equal mass, the projectile will come to rest and the target will move off with equal velocity. Hence, the velocities are interchanged i.e. the speeds are interchanged which in turn interchanges the momentum. Also, if target have some velocity then the faster body slows down and the slower body speed up.

The spring constant k = 9 N/m and mass m = 1 kg give an angular frequency omega = sqrt(k/m) = 3 rad/s. Natural length is 3m, block is released from x = 1m with natural length of spring 3m means initial stretch is 2m. Without the wall, it would oscillate as x = 3 - 2cos(3t). Because of the elastic wall at x = 3m, the motion on the right side is reflected, leading to the absolute value form x = 3 - abs(2cos(3t)).

Here, a steel ball moving with a velocity $\bar v$ collides with an identical ball originally at  rest. hence, masses of two steel balls are equal. For a head-on collision with a stationary object of equal mass, the projectile will come to rest and the target will move off with equal velocity, thus, the velocity of the first ball after the collision is zero.

In the elastic collision between a heavy object and a very light object at rest, the velocity of particles after collision is 
for heavy particle, $v _1 = u _1$
for light particle, $v _2 = 2u _1 - u _2$
since, $u _2 = 0$ hence, 
$v _2 = 2u _1$
Therefore, the stone will fly away with a velocity equal to 
$v _2 = 2u _1 = 2(10) = 20 ms^{-1}$

$\begin{array}{l} Total, , initial, , energy, , of, , two, , particles \ =\frac { 1 }{ 2 } { m _{ 1 } }{ u _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ u _{ 2 } }^{ 2 } \ Total, , final, , energy, , of, , two, particles \ =\frac { 1 }{ 2 } { m _{ 1 } }{ v _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ v _{ 2 } }^{ 2 }+\in  \ U\sin  g, , energy, , conservation, , principle, \ \frac { 1 }{ 2 } { m _{ 1 } }{ u _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ u _{ 2 } }^{ 2 }=\frac { 1 }{ 2 } { m _{ 1 } }{ v _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ v _{ 2 } }^{ 2 }+\in  \ \therefore \frac { 1 }{ 2 } { m _{ 1 } }{ u _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ u _{ 2 } }^{ 2 }-\in =\frac { 1 }{ 2 } { m _{ 1 } }{ v _{ 1 } }^{ 2 }+\frac { 1 }{ 2 } { m _{ 2 } }{ v _{ 2 } }^{ 2 } \end{array}$

For an elastic collision between two equal masses where one is initially at rest, the angle between the final velocity vectors is always 90 degrees (pi/2).

The time taken to reach the highest point after the rebound at speed v = 10 m/s is t_up = v/g = 10/10 = 1 s. However, the ceiling is at a height of 3.2 m. Using kinematic equation y = vt - (1/2)gt^2, at y = 3.2m we get 3.2 = 10t - 5*t^2, which solves to t = 0.4 s (the ball hits the ceiling before reaching its natural peak of 5m). The collision with the ceiling reverses the velocity, so the round trip time until it returns to the floor is 0.8 s.

Initial kinetic energy of the 3 kg ball is (1/2)3(4)^2 = 24 J. After an elastic collision, its kinetic energy is 6 J. Using the conservation of kinetic energy and momentum for an elastic collision in 1D, the final speed of the 3 kg mass can be found, leading to possible mass values of m = 1 kg or m = 9 kg.

In any isolated system undergoing an elastic collision, total kinetic energy is conserved. Thus, the decrease in kinetic energy of one body must equal the increase in kinetic energy of the other body. Since one body loses 100 J, the other body gains exactly 100 J, regardless of their mass ratio.

When two identical masses collide elastically in one dimension, they simply exchange their velocities. Since A starts at 0.5 m/s and B at -0.3 m/s, after the collision, A will have -0.3 m/s and B will have 0.5 m/s.

For a head-on elastic collision where a very massive body (m1) hits a stationary light body (m2), the velocity of the light body after the collision is twice the velocity of the incident massive body.

A ball falling freely from height h takes time t = sqrt(2h/g) to reach the floor. For a perfectly elastic collision, it rebounds with the same speed and takes the same time t to return to the original height. The total time elapsed before it hits the floor again is t' = 2t, making the ratio t' to t equal to 2:1.

In a perfectly elastic collision between two identical masses where one is initially at rest, the two bodies will move at an angle of 90 degrees to each other after the collision, provided the collision is oblique.

If the kinetic energy of the neutron is less than the excitation energy of the hydrogen atom (13.6 eV), the collision cannot result in internal energy changes, so it must be elastic to conserve energy and momentum.

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.

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

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.

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

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

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.

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.

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

From conservation of linear momentum:

When bodies of same mass collide head on elastically then after collision they exchange their velocities.

Here,  ${m} _{1} = {m} _{2} = m$,     ${u} _{1} = u$,    ${u} _{2} = 0$
Let ${v} _{1},   {v} _{2}$ be their velocities after collision.
According to principle of conservation of linear momentum
$mu + 0 = m\left({v} _{1}+{v} _{2}\right)$
or   ${v} _{1} + {v} _{2} = v$       ....(i)
By definition,   $e = \dfrac{{v} _{2} - {v} _{1}}{u-0}$
or   ${v} _{2} - {v} _{1} = eu$      .....(ii)
Adding equations (i) and (ii), we get
${v} _{2} = \dfrac{u\left(1+e\right)}{2}    \Rightarrow    \dfrac{{v} _{2}}{u} = \dfrac{1+e}{2}$

Mass = $0.06kg$

Since the collision is elastic, we know that when two bodies of same mass have an elastic collision their velocities get interchanged.
$m _1u _1+m _2u _2=m _1v _1+m _2v _2$

Since the body is much heavy these won't be much change in velocity
& $e = 1$

A : Only possible case.
B : Since masses are not same velocities can't be same.

For an elastic collision $($ using momentum conservation and coefficient of restitiution $)$,

Coefficient of restitution is given by,

Make your co-ordinate axes such that 1 is along the line joining the centres of the 2 just at time on collision and one, tangential. Now consider the velocity of the coming ball in 2 components of the coordinate axes we just assumed, say vn and vt(v-normal and v-tangential). Now during collision, v-tangential can't change as there is no impact along this direction. And collision along normal is just like collision in 1-D. The equations are exactly same. Hence the 1st particle will have 0 velocity along normal and second will move along normal. Hence, they'll move perpendicular to each other. 

$m(12)+m(0)=m{ v } _{ 1 }+m{ v } _{ 2 }=0\Rightarrow { v } _{ 1 }+{ v } _{ 2 }=12\ $
$-\dfrac { { v } _{ 1 }-{ v } _{ 2 } }{ 12-(0) } =1\Rightarrow { v } _{ 1 }-{ v } _{ 2 }=-12\ $
$thus,\quad { v } _{ 1 }=0\ { v } _{ 2 }=12,\quad so\quad velocity\quad of\quad first\quad after\quad collision\quad is\quad 0.$

The velocity at the top of a circle of radius l is sqrt(gl). For an elastic collision between equal masses, they exchange velocities. The second bob must acquire sqrt(gl2) to complete the circle, and the first bob must have had at least sqrt(gl1) to reach the top. Equating the velocities leads to the ratio.

In a series of collisions, the velocity of the nth ball is multiplied by a factor related to the coefficient of restitution e for each collision.

The principle of conservation of linear momentum can be strictly applied during a collision between two particles provided the time of impact is  $Moderately,,small.$