Q.1 -> If a particle of mass m is dropped from a height (say h) and after collision with  the floor it comes to rest. What is the force exerted by the particle of mass m to the floor?

Asked by BIPLAV BIJOY GOSWAMI | 23rd Mar, 2014, 08:08: PM

Expert Answer:

The particle collides with the floor for a time interval say Δt and comes to rest.
 
The acceleration of the particle is
 
straight a equals fraction numerator increment straight v over denominator increment straight t end fraction equals fraction numerator 0 minus straight u over denominator increment straight t end fraction equals fraction numerator minus straight u over denominator increment straight t end fraction
Thus, the magnitude of force is
 
straight F equals ma equals fraction numerator mu over denominator increment straight t end fraction
Now, the speed is given as
 
straight u equals square root of 2 gh end root
Hence, the force of particle on the floor is
 
straight F equals fraction numerator straight m square root of 2 gh end root over denominator increment straight t end fraction

Answered by  | 14th Apr, 2014, 03:35: PM

Queries asked on Sunday & after 7pm from Monday to Saturday will be answered after 12pm the next working day.