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

In one - dimensional head on collision, the relative velocity of approach before collision is equal to :

  1. relative velocity of separation after collision

  2. $e$ times relative velocity of separation after collision

  3. $1/e$ times relative velocity of separation after collision

  4. sum of the velocities after collision
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Coefficient of restitution is given by,

$e = \dfrac{\text{velocity of separation along line of impact}}{\text{Velocity of approach line of impact}}$

$\therefore e = \dfrac{v _2-v _1}{u _1- u _2}$

$\therefore u-1 - u _2 = \dfrac{1}{e} (v _2-v _1)$

$\therefore$ option C is correct.

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

A body of mass 'm' moving with certain velocity collides with another identical body at rest. If the collision is perfectly elastic and after the collision, if both the bodies move, they can move

  1. in the same direction

  2. in opposite direction

  3. in perpendicular direction

  4. making $45^o$ to each other
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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. 

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

 A marble going at a speed of $12 ms^{-1}$ hits another marble of equal mass at rest. If the collision is perfectly elastic. Find the velocity of the first after collision.

  1. <span>$4$</span>

  2. $0$

  3. <span>$2$</span>

  4. <span>$3$</span>
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$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\ $
$thus,\quad { v } _{ 1 }=0\ { v } _{ 2 }=12,\quad so\quad velocity\quad of\quad first\quad after\quad collision\quad is\quad 0.$

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

A ball is dropped from height h on the ground. If the coefficient of restitution is e, the height to which the ball goes up after it rebounds for the nth time is :

  1. $\dfrac{h}{e^{2n}}$

  2. $\dfrac{e^{2n}}{h}$

  3. $he^{2n}$

  4. $he^n$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let $v _0$ be the velocity with which the ball strikes the Earth first time and $v _n$ after the nth rebound.
The coefficient of restitution is
$e=\dfrac{v _1}{v _0}=\dfrac{v _2}{v _1}=\dfrac{v _3}{v _2}=$.....$=\dfrac{v _n}{v _{n-1}}$
$\therefore e^n=\dfrac{v _1}{v _0}\times \dfrac{v _2}{v _1}\times\dfrac{v _3}{v _2}\times $.....$\times \dfrac{v _n}{v _{n-1}}=\dfrac{v _n}{v _0}$
Here, $v _0=\sqrt{2gh}$ and $v _n=\sqrt{2gH}$
$e^n=\dfrac{\sqrt{2gH}}{\sqrt{2gh}}=\dfrac{\sqrt{H}}{\sqrt{h}}$
$e^{2n}=\dfrac{H}{h}\Rightarrow H=he^{2n}$
Multiple choice collisions in one dimension collisions work, energy and power mechanics physics

State whether the following statements are true or false with reasons.
Elastic forces are always conservative.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

False, because elastic forces are not always conservative. Elastic forces are conservative as long as the loading and deloading curves are coincident, even if the curves are not linear. For example, rubber.

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

A bob of mass m, suspended by a string of length $l _1$ is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point,  it collides elastically with another bob of mass m  suspended by a string of length $l _2$, which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the ratio $l _1 / l _2$ is

  1. 12

  2. 5

  3. 3

  4. 2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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(g*l2) to complete the circle, and the first bob must have had at least sqrt(g*l1) to reach the top. Equating the velocities leads to the ratio.

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

A gun is mounted on a trolley free to move on horizontal tracks. Mass of gun and trolley $25$ kg. Gun fires a two shells of mass $5$ kg each other. Velocity of shells with roll to gun is $60 $ m/s . In above question, velocity of gun with respect to ground after firing second shell is-

  1. $12$ m/s

  2. $18$ m/s

  3. $50$ m/s

  4. $60$ m/s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using conservation of momentum for the system (gun + trolley + shells). Total mass = 25 kg. Each shell = 5 kg. Velocity of shell relative to gun = 60 m/s. This is a multi-step recoil problem.

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

A gun is mounted on a railroad car. The mass of the car, the gun, the shells and the operator is $50$m where m is the mass of the one shell. If the muzzle velocity of shell is $200$ m/s, what is recoil speed of car after second shot?

  1. $\dfrac{200}{49}$ m/s

  2. $200\left(\dfrac{1}{48}+\dfrac{1}{48}\right)$ m/s

  3. $200\left(\dfrac{1}{48}+\dfrac{1}{49}\right)$ m/s

  4. $200\left(\dfrac{1}{48}+\dfrac{1}{48\times 49}\right)$ m/s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Centre of mass of velocity of shell = 0 (initially at rest)

Therefore,$ 0=49  m\times v+m\times200$

$v=\dfrac{-200}{49}m/s$

$\dfrac{200}{49}m/s$

Another shell is fire , then the velocity of the car , with respect to platform is

$v'=\dfrac{200}{49}m\ s$ toward left

Another shell is fired , then the velocity of the car with respect to platform is

$v'=\dfrac{200}{48}m\ s$ towards left



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

N identical balls are placed on a smooth horizontal surface. Another ball of same mass collides elastically with velocity $u$ with first ball of N balls. A process of collision is thus started in which first ball collides with second ball and the second ball with the third ball and so on. The coefficient of resulting for each collision is $e$. Find speed of Nth ball :

  1. $(1+e)^{N}u$

  2. $u(1+e)^{N-1}$

  3. $\cfrac{u(1+e)^{N-1}}{2^{N-1}}$

  4. $u^{N}(1+e)^{N-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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.

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

A neutron moving with velocity u collides with a stationary $\alpha -particle$ The velocity of the neutron after collision is 

  1. $-\dfrac { 3\cup }{ 5 } $

  2. $\dfrac { 3\cup }{ 5 } $

  3. $\dfrac { 2\cup }{ 5 } $

  4. $-\dfrac { 2\cup }{ 5 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Mass of alpha particle is $4$ times to neutron 
according to the condition alpha paticle is at rest so its velocity is zero
now the final velocity of neutron is calculated from this formula
$U' = (m-1-m _2÷ m _1+m _2) U^1$
by putting$,$ 
$U' = (1-4÷1+4) $
$= -3U/5 $
Negative sign shows reverse direction of neutro$.$
Hence,
option $(A)$ is correct answer.