Filters

Clear All
By Einstein’s mass-energy relation we can write :                                                  Here  and C are constant thus two-body decay is unable to explain (or account for) the continuous energy distribution in the β-decay of a neutron.
The potential energy of the system depends inversely on the separation between the balls. Thus the potential energy will decrease as the balls will come closer and will become zero as they touch each other. Thus elastic collision is best described only by the graph (v).
The initial momentum of the system (boy + trolley) is given as :                                                                                                                                            Now assume v' is the final velocity of the trolley with respect to the ground. Then the final momentum will be :                                                         Conserving momentum :  ...
In this case, the heat produced is the loss in the potential energy.  Thus,            heat produced  =  mg h or                                            or                                            The heat produced (when the lift is stationary) will remain the same as the relative velocity of the bolt with respect lift still remains zero.
Displacement (x) of the block is given as :   =  0.1 m. Using equilibrium conditions we can write :                                                 and                                                                           (  is the frictional force). We can write work done in terms of potential energy as :                                            or                                     ...
The FBD of the track is shown in the figure below :                                                Using the law of conservation of energy we have :                                             or                                             Hence both stones will reach the bottom with the same speed. For stone 1 we can write :                                     or                               ...
We are given :                       Mass of the bullet m: 0.012 Kg      Mass of the block M: 0.4  Kg   The initial velocity of the bullet u: 70  m/s  The initial velocity of the block     :     0   The final velocity of the system (bullet + block): v For finding the final speed of system we will apply the law of conservation of momentum :                                     or                 ...
Exams
Articles
Questions