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A ball of mass 2kg, charge 1×10-6 C is dropped from top of a high tower . In space electric field exists in horizontal direction away from the tower whoch vareis as E=(5-2x)×106 V/m. Find the maximum horizontal distance ball can go from the tower and displacement of ball at this instant from the point of projection?
hi , look the figure ,below . it shows only how the horizonta motion occurs dv/dt = (5-2x)×106 ( 1×10-6 ) =(5-2x) dx/dt = v dv/dx = (5-2x)/v v dv = (5-2x) dx v^2/ 2] from 0 to v = 5x - x^2 ] from o to x v^2/ 2 = 5x - x^2 for v =0 , 5x - x^2 = 0 x= 0, 5 so ball will move back at x= 5 , this will be max. horizontal distance displacement of ball at this instant from the point of projection , dx/ dt = v = √ 10x - 2 x^2 from here we can get time when it reaches x= 5 , in this duration vertical displacement = 1/2 g t ^2 total displacement = √ ( horiz. displacemen ) ^2 + ( vertical displacement ) ^2
hi ,
look the figure ,below . it shows only how the horizonta motion occurs
dv/dt = (5-2x)×106 ( 1×10-6 )
=(5-2x)
dx/dt = v
dv/dx = (5-2x)/v
v dv = (5-2x) dx
v^2/ 2] from 0 to v = 5x - x^2 ] from o to x
v^2/ 2 = 5x - x^2
for v =0 ,
5x - x^2 = 0
x= 0, 5
so ball will move back at x= 5 , this will be max. horizontal distance
displacement of ball at this instant from the point of projection ,
dx/ dt = v = √ 10x - 2 x^2
from here we can get time when it reaches x= 5 ,
in this duration vertical displacement = 1/2 g t ^2
total displacement = √ ( horiz. displacemen ) ^2 + ( vertical displacement ) ^2
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