Tag: work, energy and power

Questions Related to work, energy and power

A gun fires a shell of mass $1.5 $kg with velocity of $150$ m/s and recoils with a velocity of $2.5$ m/s. 

Calculate the mass of the gun.

  1. $20$ kg

  2. $30$ kg

  3. $90$ kg

  4. $60$ kg


Correct Option: C
Explanation:

Mass of shell $=1.5$kg

Velocity of shell ${V _1} = 159\,m/s$
Recoil velocity $=2.5$ m/s
mass of gun $=M$
According to the conservation of momentum,
$m{v _1} = m{v _2}$
$\begin{array}{l} 1.5\times 150=M\times 2.5 \ M=\dfrac { { 1.5\times 159 } }{ { 2.5 } }  \ M=90\, kg \end{array}$

A particle of mass m, is attached to one end of a massless spring of force constant k, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time t = 0, with an initial velocity $u _0$. When the speed of the particle is $0.5\, u _0$, it collides elastically with a rigid walk. After this collision:

  1. The speed of the particle when it returns to its equilibrium position is $u _0$

  2. The time at which the particle passes through the equilibrium position for the first time is $t = \pi \sqrt{\dfrac{m}{k}}$

  3. The time at which the maximum compression of the spring occurs is $t = \dfrac{4\pi}{3}\sqrt{\dfrac{m}{k}}$

  4. The time at which the particle passes througout the equilibrium position for the second time is $t= \dfrac{4\pi}{3}\sqrt{\dfrac{m}{k}}$


Correct Option: A

A ball hits the floor and rebounds after an elastic collision. In this case:

  1. the momentum of the ball just after the collision is same as that just before the collision.

  2. the kinetic energy of the ball remains same during collision.

  3. the total momentum of the ball and the earth is conserved.

  4. the total energy of the ball and the earth remains the same.


Correct Option: C,D
Explanation:

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


Since the collision is inelastic a part of the mechanical energy is lost hence (b) is not true.  

Taking earth and the ball as a system there is no external force on the system. Hence the total momentum of the ball and the earth is conserved. So (c) is true.  

From the conservation principle of the energy, the total energy of the ball and the earth remains the same. Hence (d) is true.

Two identical spheres move in opposite direction with speed $v _1$ and $v _2$ and pass behind an opaque screen, where they may either cross without touching ( Event 1) or make an elastic head-on collision ( Event 2)

  1. We can never make out which events has occured

  2. We cannot make out which event has occured only if $v _1= v _2$

  3. We can always make out which event has occured

  4. We can make out which event has occured only if


Correct Option: A

A bullet of mass m is fired into a large block of wood of mass M with velocity v. The final velocity of the system is?

  1. $\left(\dfrac{m}{M-m}\right)v$

  2. $\left(\dfrac{m+M}{M}\right)v$

  3. $\left(\dfrac{M-m}{M}\right)v$

  4. $\left(\dfrac{m}{m+M}\right)v$


Correct Option: D
Explanation:

u is initial velocity 

v is finall velocity 
m1 is m 
m2 is M 
u1 is v 
u2 is 0 
${ v } _{ 1 }={ v } _{ 2 }={ v } _{ 3 }$
$total\quad initial\quad momentum=finall\quad momentum$
${ m } _{ 1 }{ u } _{ 1 }+{ m } _{ 2 }{ u } _{ 2 }={ m } _{ 1 }{ v } _{ 1 }+{ m } _{ 2 }{ v } _{ 2 }$
$mv+M\times 0=m{ v } _{ 3 }+{ Mv } _{ 3 }$
$mv=\left( m+M \right) { v } _{ 3 }$
${ v } _{ 3 }=\dfrac { mv\quad  }{ m+M } $

sphere collides with another sphere of identical mass kept at rest. Mier collision, the two spheres move. The collision is perfectly inelastic, then the angle between the directions of motion of the two spheres is

  1. ${ 0 }^{ o }$

  2. ${ 45 }^{ o }$

  3. different from ${ 90 }^{ o }$

  4. ${ 90 }^{ o }$


Correct Option: A

A smooth sphere A of mass m collides elastically with an identical sphere B at rest. The velocity of A before collision is $8 m/s$ in a direction making $60^{o}$ with the line joining the centres at the time of impact. Which of the following is/are possible:

  1. the sphere a comes to rest after collision

  2. the sphere B will move with a speed of $8 m/s$ after collision

  3. the directions of motion of A and B after collision are at right angles

  4. the speed of B after collision is $2 m/s$


Correct Option: C

An object is moving towards a mirror with  a velocity v as shown in figure.if the collision between the mirror and the object is perfectly elastic, then the velocity of the image after collision with mirror in vector form is.

  1. $ - v\;\widehat j$

  2. $ - v\;\cos \;2\theta \widehat j + v\;\sin \;2\theta \;\widehat i$

  3. $ - v\widehat i$

  4. $ - v\;\cos \theta \;j - \;v\;\sin \theta \;\widehat i$


Correct Option: A

A body of mass $m$ moving with velocity $u$ collides elastically with another body of mass $m$ at rest. After collision, they moves in a plane with velocities $V _{1}$ and $V _{2}$ respectively. Then the angle beteen $\vec{V} _{1}$ and $\vec{V} _{2}$ is

  1. $zero$

  2. $60^{o}$

  3. $90^{o}$

  4. $180^{o}$


Correct Option: D

Two solid balls of rubber $A$ and $B$ whose masses are $200\ gm$ and $400\ gm$ respectively, are moving in mutually opposite directions. if the velocity of ball A is $0.3\ m/s$ and both the ball come to rest after collision, then the velocity of ball $B$ is :

  1. $0.15\ m/s$

  2. $-0.15\ m/s$

  3. $1.5\ m/s$

  4. $None\ of\ these$


Correct Option: B
Explanation:
Initial linear momentum of system 

$= [{{m} _{A}}{{\vec{v}} _{A}}+{{m} _{B}}{{\vec{v}} _{B}}]$

$= 0.2 \times  0.3 + 0.4 \times  vB$

Finally both balls come to rest \ final linear momentum = 0 

By the law of conservation of linear momenum        

$0.2 \times  0.3 + 0.4 \times  vB = 0$

$[{{v} _{B}}=-\dfrac{0.2\times 0.3}{0.4}=-0.15\ m/s]$