
A particle is projected with a velocity $u$ in horizontal direction as shown in the figure. find $u$(approx.) so that the particle collides orthogonally with the inclined plane of the fixed wedge.

A) $10m/s$
B) $20m/s$
C) $10\sqrt 2 m/s$
D) None of these
Answer
133.5k+ views
Hint: In order to solve this question you have to break the velocity components of the moving particle and write the force equation and then apply the equilibrium condition. You should remember all the concepts related to wedges. Also, keep in mind that in this question the particle’s velocity component parallel to the inclined plane should be zero.
Complete step by step solution:
Considering plane along the wedge axis,
As we know that when a particle collides with the plane, its velocity component which is parallel to the inclined plane should be zero.
$ \Rightarrow u\cos \theta - g\sin \theta \times t = 0$
On further solving, we have
$ \Rightarrow t = \dfrac{u}{{g\tan \theta }}$
And also from the same equation, we get the value of u as
$ \Rightarrow u = gt \times \tan \theta $
On putting $\theta = 30^\circ $ in the above equation, we have
$ \Rightarrow u = \dfrac{{gt}}{{\sqrt 3 }}$ …….(1)
Now, consider the plane along with the x-y axis, we have
Horizontal displacement before the particle hits the plane $ = ut$
And, the vertical displacement during the t time $ = \dfrac{1}{2}g{t^2}$
Now, $\tan 30^\circ = \dfrac{{h - \dfrac{1}{2}g{t^2}}}{{ut}}$
$ \Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{{h - \dfrac{1}{2}g{t^2}}}{{ut}}$
On solving the above equation for t, we have
$ \Rightarrow t = \sqrt {\dfrac{{6h}}{{5g}}} $
On putting the given values $h = 55m$ and $g = 9.8m/{s^2}$ we get
$ \Rightarrow t = 2.5938\sec $
Now, putting that above value of time in the equation (1), we have
$ \Rightarrow u = \dfrac{{9.8 \times 2.5938}}{{\sqrt 3 }}$
On further solving this, we get the value of the initial velocity $u$ ,
$ \Rightarrow u = 14.69 \simeq 10\sqrt 2 m/s$
Therefore, the correct option is (C).
Note: Always keep in mind that if the wedge is at rest then the problem should be solved by the normal way in which we write the force equations and applying the equilibrium conditions and if the wedge is in accelerating motion then everything will be converted into a wedge frame of reference and then solve the question by applying pseudo force.
Complete step by step solution:
Considering plane along the wedge axis,
As we know that when a particle collides with the plane, its velocity component which is parallel to the inclined plane should be zero.
$ \Rightarrow u\cos \theta - g\sin \theta \times t = 0$
On further solving, we have
$ \Rightarrow t = \dfrac{u}{{g\tan \theta }}$
And also from the same equation, we get the value of u as
$ \Rightarrow u = gt \times \tan \theta $
On putting $\theta = 30^\circ $ in the above equation, we have
$ \Rightarrow u = \dfrac{{gt}}{{\sqrt 3 }}$ …….(1)
Now, consider the plane along with the x-y axis, we have
Horizontal displacement before the particle hits the plane $ = ut$
And, the vertical displacement during the t time $ = \dfrac{1}{2}g{t^2}$
Now, $\tan 30^\circ = \dfrac{{h - \dfrac{1}{2}g{t^2}}}{{ut}}$
$ \Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{{h - \dfrac{1}{2}g{t^2}}}{{ut}}$
On solving the above equation for t, we have
$ \Rightarrow t = \sqrt {\dfrac{{6h}}{{5g}}} $
On putting the given values $h = 55m$ and $g = 9.8m/{s^2}$ we get
$ \Rightarrow t = 2.5938\sec $
Now, putting that above value of time in the equation (1), we have
$ \Rightarrow u = \dfrac{{9.8 \times 2.5938}}{{\sqrt 3 }}$
On further solving this, we get the value of the initial velocity $u$ ,
$ \Rightarrow u = 14.69 \simeq 10\sqrt 2 m/s$
Therefore, the correct option is (C).
Note: Always keep in mind that if the wedge is at rest then the problem should be solved by the normal way in which we write the force equations and applying the equilibrium conditions and if the wedge is in accelerating motion then everything will be converted into a wedge frame of reference and then solve the question by applying pseudo force.
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Class 11 JEE Main Physics Mock Test 2025

Current Loop as Magnetic Dipole and Its Derivation for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Motion In A Plane: Line Class 11 Notes: CBSE Physics Chapter 3

Waves Class 11 Notes: CBSE Physics Chapter 14
