Laws Of Motion Question 95

Question: A particle of mass m, initially at rest, is acted upon by a variable force F for a brief interval of time T. It begins to move with a velocity u after the force stops acting. F is shown in the graph as a function of time. The curve is a semicircle.

Options:

A) $ u=\frac{\pi F _{0}^{2}}{2m} $

B)$ u=\frac{\pi T^{2}}{8m} $

C) $ u=\frac{\pi F _{0}T}{4m} $

D)$ u=\frac{F _{0}T}{2m} $

Show Answer

Answer:

Correct Answer: C

Solution:

Initially particle was at rest. By the application of force its momentum increases.

Final momentum of the particle $ = $ Area of F - t graph

therefore $ mu= $ Area of semicircle $ mu=\frac{\pi \ r^{2}}{2} $

$ =\frac{\pi \ r _{1}r _{2}}{2} $

$ =\frac{\pi \ (F _{0})\ (T/2)}{2} $ therefore$ u=\frac{\pi \ F _{0}T}{4m} $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक