AP EAMCET · PHYSICS · Motion In One Dimension
At the moment \(\mathrm{t}=0\), a time dependent force \(\mathrm{F}=\) at (where a is constant equal to \(1 \mathrm{Ns}^{-1}\) ) is applied to a body of mass \(1 \mathrm{~kg}\) resting on a smooth horizontal plane as shown in the figure. If the direction of this force makes an angle \(45^{\circ}\) with the horizontal, then the velocity of the body at the moment it leaves the plane is
(acceleration due to gravity \(=10 \mathrm{~m} \mathrm{~s}^{-2}\) )

- A \(50 \mathrm{~ms}^{-1}\)
- B \(50 \sqrt{2} \mathrm{~ms}^{-1}\)
- C \(100 \sqrt{2} \mathrm{~ms}^{-1}\)
- D \(100 \mathrm{~ms}^{-1}\)
Answer & Solution
Correct Answer
(B) \(50 \sqrt{2} \mathrm{~ms}^{-1}\)
Step-by-step Solution
Detailed explanation
When body left the surface, \(\mathrm{N}=0\) So, \(m g=\) at \(\sin 45^{\circ}\) \(\mathrm{t}=\frac{\mathrm{mg}}{\mathrm{a} \sin 45^{\circ}}=\frac{10}{\frac{1.1}{\sqrt{2}}}=10 \sqrt{2} \mathrm{sec}\) Now,…
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