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} $
BETA
