Question

A small object of mass m and of charge q slides down along a frictionless slope of a elevation $\theta$ from a height h. From the midpoint of the slope, it enters into a region of uniform horizontal electric field, and it decelerates and stops at the bottom of the slope.

• A. The magnitude of E is $\frac{mg}{q}tan\alpha$
  
• B. The normal reaction on block before dotted line and after dotted line remains the same
• C. The magnitude of acceleration before and after dotted line remains the same
• D. The speed at height $\frac{h}{4}$ from bottom is $\sqrt{\frac{gh}{2}}$

Answer 1

Answer: (C), (D)

The speed at height $\frac{h}{4}$ from bottom is $\sqrt{\frac{gh}{2}}$

$v^2=0^2+2g\sin \theta x$

$0^2=v^2-2ax$
$a=g\sin \theta =\frac{qE\cos \theta -mg\sin \theta }{m}$
$E=\frac{2mg\sin \theta }{q\cos \theta }=\frac{2mg\tan \theta }{q}$
$\frac{1}{2}m{v}^2=mg\times \frac{h}{4}$
$v^2=\frac{gh}{2}$
$v=\sqrt{\frac{gh}{2}}$

The speed at height $\frac{h}{4}$ from bottom is same as that at height $\frac{h}{4}$ from top.