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Two particles move at right angle to each other de Broglie wavelengths as \lambda_1 and \lambda_2, particle suffers a perfectly inelastic collision. The de Broglie wavelength \lambda of final particle is given by

  • Option 1)

    \lambda=\frac{\lambda _{1}\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

  • Option 2)

    \lambda=2\frac{\lambda _{1}\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

  • Option 3)

    \lambda=\frac{\lambda _{1}^2\lambda _{2}}{\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

  • Option 4)

    \lambda=\frac{\lambda _{1}\lambda _{2}}{2\sqrt{\lambda_{1}^{2}+\lambda_{2}^{2}}}

 

option 3

 

Two particles , of masses M and 2M , moving as shown , with speeds of 10 m/s and 5 m/s ,

collide elastically at the origin.After the collision , they move along the indicated directions with

speeds v_{1} and v_{2}, respectively. The values of v_{1} and v_{2} are nearly :

  • Option 1)

    6.5 m/s and 6.3 m/s

  • Option 2)

    3.2 m/s and 6.3 m/s

  • Option 3)

    6.5 m/s and 3.2 m/s

  • Option 4)

    3.2 m/s and 12.6 m/s

 
  Elastic Collision in 2 dimension - - wherein                                                                                                                                                                                                                                                                      ................................(1)                  ...

A body of mass m_{1} moving with an unknowm velocity of v_{1}\hat{i}, undergoes a collinear collision with a body of mass m_{2} moving with a velocity v_{2}\hat{i}. After collision ,m_{1} and m_{2} move with velocities of v_{3}\hat{i} and v_{4}\hat{i}, respectively.

If m_{2}=0.5m_{1} and v_{3}=0.5v_{1}, the v_{1} is :

 

  • Option 1)

    v_{4}-\frac{v_{2}}{2}

  • Option 2)

    v_{4}-v_{2}

  • Option 3)

    v_{4}-\frac{v_{2}}{4}

  • Option 4)

    v_{4}+v_{2}

before collision: after collision:   &   Option 1)Option 2)Option 3)Option 4)

A particle of mass 'm'  is moving with speed '2 v' and collides with a mass '2 m' moving with speed 'v' in the same direction. After collision , the first mass is stoped completely while the second one splits into two particles each of mass 'm', which move at angle 45^{0} with respect to the original direction.

The speed of each of the moving particle will be :

  • Option 1)

    \sqrt{2}v

  • Option 2)

    2\sqrt{2}v

  • Option 3)

    v/(2\sqrt{2})

  • Option 4)

    v/\sqrt{2}

 
   in  direction Option 1) Option 2) Option 3) Option 4)

Two particles move at right angle to each other. Their de Broglie wavelengths are \lambda _{1} and \lambda _{2} respectively. The particles suffer perfectly inelastic collision. The de Broglie wavelength \lambda, of the final particle, is given by :
 

  • Option 1)

    \lambda=\frac{\lambda_{1}+\lambda_{2}}{2}

  • Option 2)

    \lambda=\sqrt{\lambda_{1}+\lambda_{2}}

  • Option 3)

    \frac{1}{\lambda_{2}}=\frac{1}{\lambda_{1}^{2}}+\frac{1}{\lambda_{2}^{2}}

     

  • Option 4)

    \frac{2}{\lambda}=\frac{1}{\lambda_{1}}+\frac{1}{\lambda_{2}}

 
We know or Option 1) Option 2) Option 3)   Option 4)

A body of mass 2 \; kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed.What is the mass of the second body ?

  • Option 1)

    1.0 \: kg

  • Option 2)

    1.5\: kg

  • Option 3)

     1.8\: kg

  • Option 4)

    1.2 \: kg

 
 mass of  body =   initial velocity of  body   mass of  body A                      B                                                           rest       (from e=1) momentum balance                             Option 1) Option 2) Option 3)   Option 4)

A particle moves in one dimension from rest under the influence of a force that varies with the distance travelled by the particle as shown in the figure. The kinetic energy of the particle after it has travelled 3 m is :

  • Option 1)

    6.5 J
     

  • Option 2)

    4 J

  • Option 3)

    2.5 J

     

  • Option 4)

    5 J

Work Energy theorem Now work is Area of F vs distance curve. Option 1)  Option 2)Option 3)  Option 4)

A uniform cable of mass 'M' and length 'L' is placed on a horizontal surface that its \left [ \frac{1}{n} \right ]^{th}part is hanging below the edge of the surface. To lift the hanging part of the cable upto the srface, the work done should be :

  • Option 1)

      \frac{MgL}{2n^{2}}

  • Option 2)

      \frac{MgL}{n^{2}}

  • Option 3)

     \frac{2MgL}{n^{2}}

  • Option 4)

     nMgL

 
Total length of rod= Length hanging= initial P.E =  final P.E=                                                      Option 1)    Option 2)    Option 3)   Option 4)  
Work done by frictional is zero when the object to which it is acing does not move any distance.

Arrangement of two block system is as shown in the figure . mass of block A is m _A = 5 Kg 

and mass of B is m _B = 10 Kg 

A constant force F = 100 N applied on upper block A . Friction between A and B is \mu . and b/w B and ground surface is smooth , then find work done by frictional force on block B in t = 2 sec  if system starts from rest 

friction force on block A,  An equal and opposite force will be applied on block B Work done = 
A lift of mass m is moving upwards direction with acceleration g/4 , its displacement is n. find work done by tension force.
apply force balance  now work done by T = 

A mass m slides from rest at height h down a curved surface which become horizontal at zero height (see figure). A sping is fixed horizontally on the level part of the surface. The spring constant is K N/m. When the mass encounters the spring it encounters, it compresses the spring x=h/10. If m=1Kg,h=5m find k/100

When the mass encounters the spring it encounters, it compresses the spring x=h/10. If m=1Kg,h=5m then value of  k/100 is: 
For two particles A and B, given that r(A)=2i+3j, r(B)=6i+7j, v(A)=3i-j and v(B)=xi-5j. what is the value of x if they collide.
If two particle A & B collide then the value of  x  is: 
A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg . What is the work done in pulling the entire chain on the table ?
Consider a small part dx at a depth x from table.  Work done in lifting this small portion is dw = dm gx Total work done         
Sir 6th ano 7th question please Track OABCD (as shown in figure) is smooth and fixed in vertical plane. What minimum speed has to be given to a particle lying at point A so that it can reach point C

The speed to reach B is obtained by

\frac{1}{2}mu^2=mgh\ u = initial\ velocity

u=80m/s. A slight push is enough for ball to reach C

so minimum speed is 80m/s

mechanical_energy.PNG A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated How much fat will he use up considering the work done only when the weight is lifted up?
 Total work done by the person in lifting the weigh = mgh           Total mechanical energy produced by burning 1 kg fat    Total fat burn 
A force acts on a 2kg object so that its position is given as a function of time as x = 3t2 + 5. What is the work done by this force in first 5 seconds?
Work done = change in kinetic energy                                                         

A block of mass m = 5 Kg is moving up on inclined plane of angle theta = 30 degree with horizontal with constant velocity V by the help of force F . What is the work done by force F to move the object up the inclined by distance 's' [ given mu_k = 1/ root3] b/w surface and block and rho = 2 m g = 10 m/s ^2

  be balancing force along inclined plane 

A block of mass 2 kg slides on a horizontal floor with velocity 4 m/s. It hits an uncompressed spring and compresses it till its final velocity reduces to zero the kinetic frictional force is 15 N and spring constant is 10000 N/m. Find the spring compression

Change in Kinetic energy = Work done by all forces. Here   Wf = Work done by spring  Wf = Work done by friction      On solving it x = 5.5 cm
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