Tag: work, energy and power

Questions Related to work, energy and power

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


Correct Option: A

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


Correct Option: A
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



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


Correct Option: B

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 } $


Correct Option: A
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.

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

  1. Extremely small

  2. Moderately small

  3. Extremely large

  4. Depends on a particular case


Correct Option: B
Explanation:

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

Hence,
option $(B)$ is correct answer.