NEET

Velocity  (1) Instantaneous velocity v: In projectile motion, vertical component of velocity changes but horizontal component of velocity remains always constant. Example: When a man jumps over the hurdle leaving behind its skateboard then vertical component of his velocity is changing, but not the horizontal component, which matches with the skateboard velocity. As a result, the skateboard stays underneath him, allowing him to land on it.             Let \[{{v}_{i}}\] be the instantaneous velocity of projectile at time t direction of this velocity is along the tangent to the trajectory at point P. \[{{\vec{v}}_{i}}={{v}_{x}}i+{{v}_{y}}\hat{j}\Rightarrow {{v}_{i}}=\sqrt{v_{x}^{2}+v_{y}^{2}}=\sqrt{{{u}^{2}}{{\cos }^{2}}\,\theta +{{(u\sin \theta -gt)}^{2}}}\]                         \[{{v}_{i}}=\sqrt{{{u}^{2}}+{{g}^{2}}{{t}^{2}}-2u\,gt\sin \theta }\]      Direction of instantaneous velocity \[\tan \alpha \,=\,\frac{{{v}_{y}}}{{{v}_{x}}}=\frac{u\,\sin \,\theta -gt}{u\,\cos \theta }\]  or       \[\alpha ={{\tan }^{-1}}\left[ \tan \theta -\frac{gt}{u}\sec \theta  \right]\] (4) Change in velocity: Initial velocity (at projection point) \[{{\overrightarrow{u}}_{i}}=u\cos \theta \,\hat{i}+u\sin \theta \,\hat{j}\]                                                    Final velocity (at highest point) more...

Momentum (1) Change in momentum: Simply by the multiplication of mass in the above expression of velocity (Article-4). (i) Change in momentum (Between projection point and highest point) \[\Delta p={{\overrightarrow{p}}_{f}}-{{\overrightarrow{p}}_{i}}=-\,mu\sin \theta \,\hat{j}\] (ii) Change in momentum (For the complete projectile motion) \[\Delta p={{\overrightarrow{p}}_{f}}-{{\overrightarrow{p}}_{i}}=-\,2mu\sin \theta \,\hat{j}\] (6) Angular momentum: Angular momentum of projectile at highest point of trajectory about the point of projection is given by \[L=mvr\]          \[\left[ \text{Here }r=H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g} \right]\] \[\therefore \,\,\,\,\,\,\,L=m\,\,u\cos \theta \,\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}=\frac{m\,\,{{u}^{3}}\cos \theta {{\sin }^{2}}\theta }{2g}\] Sample problems based on momentum and angular momentum Problem 14.  A body of mass 0.5 kg is projected under gravity with a speed of 98 m/s at an angle of 30o with the horizontal. The change in momentum (in magnitude) of the body is [MP PET 1997] (a) 24.5 N?s       (b) 49.0 N?s       (c) 98.0 N?s       (d) 50.0 N?s Solution: more...

Time of Flight Time of flight: The total time taken by the projectile to go up and come down to the same level from which it was projected is called time of flight. For vertical upward motion \[0 =u\,sin\,\,\theta \,gt\Rightarrow t=\text{ }(u\,sin\,\theta /g)\] Now as time taken to go up is equal to the time taken to come down so Time of flight \[T=2t=\frac{2u\,\sin \theta }{g}\] (i) Time of flight can also be expressed as: \[T=\frac{2.{{u}_{y}}}{g}\] (where \[{{u}_{y}}\] is the vertical component of initial velocity). (ii) For complementary angles of projection \[\theta \] and \[9{{0}^{o}}\theta \] (a) Ratio of time of flight \[=\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{2u\,\sin \theta /g}{2u\,\sin (90-\theta )/g}=\tan \theta \Rightarrow \frac{{{T}_{1}}}{{{T}_{2}}}=\tan \theta \] (b) Multiplication of time of flight = \[{{T}_{1}}{{T}_{2}}=\frac{2u\sin \theta }{g}\frac{2u\cos \theta }{g}\Rightarrow {{T}_{1}}{{T}_{2}}=\frac{2R}{g}\] (iii) If \[{{t}_{1}}\] is the time taken by projectile to rise upto point p and t2 is the time taken in falling from point p more...

Horizontal Range Horizontal range: It is the horizontal distance travelled by a body during the time of flight. So by using second equation of motion             \[R=u\cos \theta \times T\]\[=\,\]\[u\cos \theta \times (2u\sin \theta /g)\]\[=\frac{{{u}^{2}}\,\sin \,2\theta }{g}\] \[R=\frac{{{u}^{2}}\,\sin \,2\theta }{g}\] (i) Range of projectile can also be expressed as: \[\operatorname{R} = u cos\,\theta \,\,\times \, T =u\cos \theta \frac{2u\sin \theta }{g}=\frac{2\,u\cos \theta \,\,u\sin \theta }{g}=\frac{\text{2}{{\text{u}}_{\text{x}}}{{u}_{y}}}{\text{g}}\]                         \ \[R=\frac{\text{2}{{\text{u}}_{\text{x}}}{{u}_{y}}}{\text{g}}\]   (where \[{{\operatorname{u}}_{x}}\,and {{u}_{y}}\] are the horizontal and vertical component of initial velocity) (ii) If angle of projection is changed from \[\theta \] to \[\theta =(90\theta )\] then range remains unchanged.             \[R'=\frac{{{u}^{2}}\sin 2\theta \,'\,}{g}=\frac{{{u}^{2}}\sin [2({{90}^{o}}-\theta )]}{g}=\frac{{{u}^{2}}\sin 2\theta \,}{g}=R\]             So a projectile has same range at angles of projection q and (90 ? q), though time of flight, maximum height and trajectories are different. These angles q and \[{{90}^{o}}\theta \] more...

Maximum Height Maximum height: It is the maximum height from the point of projection, a projectile can reach. So, by using \[{{v}^{2}}={{u}^{2}}+2as\] \[0={{(u\sin \theta )}^{2}}-2gH\] \[H=\frac{{{u}^{2}}{{\sin }^{2}}\theta }{2g}\] (i) Maximum height can also be expressed as \[H=\frac{u_{y}^{2}}{2g}\]  (where \[{{u}_{y}}\] is the vertical component of initial velocity). (ii) \[{{H}_{\max }}=\frac{{{u}^{2}}}{2g}\]  (when \[{{\sin }^{2}}\,\theta = max = 1\] i.e., \[\theta = 9{{0}^{o}}\]) i.e., for maximum height body should be projected vertically upward. So it falls back to the point of projection after reaching the maximum height. (iii) For complementary angles of projection \[\theta \,\,and\, 9{{0}^{o}}\theta \] Ratio of maximum height = \[\frac{{{H}_{1}}}{{{H}_{2}}}=\frac{{{u}^{2}}{{\sin }^{2}}\theta /2g}{{{u}^{2}}{{\sin }^{2}}({{90}^{o}}-\theta )2g}\] \[=\frac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }\] \[={{\tan }^{2}}\theta \] \[\therefore \,\,\,\,\,\frac{{{H}_{1}}}{{{H}_{2}}}={{\tan }^{2}}\theta \] Sample problem based on maximum height Problem 49. A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same effort, he throws more...

Projectile Motion on an Inclined Plane Let a particle be projected up with a speed u from an inclined plane which makes an angle \[\alpha \] with the horizontal velocity of projection makes an angle q with the inclined plane. We have taken reference x-axis in the direction of plane. Hence the component of initial velocity parallel and perpendicular to the plane are equal to \[u\cos \theta \] and \[u\sin \theta \] respectively i.e. \[{{u}_{||}}=u\cos \theta \] and \[{{u}_{\bot }}=u\sin \theta \]. The component of g along the plane is \[g\sin \alpha \] and perpendicular to the plane is \[g\cos \alpha \] as shown in the figure i.e. \[{{a}_{||}}=-g\sin \alpha \] and \[{{a}_{\bot }}=g\cos \alpha \]. Therefore the particle decelerates at a rate of \[g\sin \alpha \] as it moves from O to P. (1) Time of flight: We know for oblique projectile motion more...

Circular Motion Circular motion is another example of motion in two dimensions. To create circular motion in a body it must be given some initial velocity and a force must then act on the body which is always directed at right angles to instantaneous velocity. Since this force is always at right angles to the displacement due to the initial velocity therefore no work is done by the force on the particle. Hence, its kinetic energy and thus speed is unaffected. But due to simultaneous action of the force and the velocity the particle follows resultant path, which in this case is a circle. Circular motion can be classified into two types ? Uniform circular motion and non-uniform circular motion. Variables of Circular Motion. (1) Displacement and Distance: When particle moves in a circular path describing an angle q during time t (as shown more...

Centripetal Acceleration and Force (1) Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration. (2) It always acts on the object along the radius towards the centre of the circular path. (3) Magnitude of centripetal acceleration \[a\,=\frac{{{v}^{2}}}{r}={{\omega }^{2}}r=4\pi {{n}^{2}}r=\frac{4{{\pi }^{2}}}{{{T}^{2}}}r\] (4) Direction of centripetal acceleration: It is always the same as that of \[\Delta \overrightarrow{\upsilon }\]. When Dt decreases, Dq  also decreases. Due to which \[\Delta \vec{\upsilon }\] becomes more and more perpendicular to \[\overrightarrow{\upsilon }\]. When \[\Delta t\to 0\], \[\Delta \overrightarrow{\upsilon }\] becomes perpendicular to the velocity vector. As the velocity vector of the particle at an instant acts along the tangent to the circular path, therefore \[\Delta \overrightarrow{\upsilon }\] and hence the centripetal acceleration vector acts along the radius of the circular path at that point and is directed towards the centre of the circular path.             Sample more...

Centrifugal Force It is an imaginary force due to incorporated effects of inertia. When a body is rotating in a circular path and the centripetal force vanishes, the body would leave the circular path. To an observer A who is not sharing the motion along the circular path, the body appears to fly off tangential at the point of release. To another observer B, who is sharing the motion along the circular path (i.e., the observer B is also rotating with the body with the same velocity), the body appears to be stationary before it is released. When the body is released, it appears to B, as if it has been thrown off along the radius away from the centre by some force. In reality no force is actually seen to act on the body. In absence of any real force the body tends to continue its motion in a more...

Work Done by Centripetal Force The work done by centripetal force is always zero as it is perpendicular to velocity and hence instantaneous displacement. Work done = Increment in kinetic energy of revolving body Work done = 0 Also \[W=\overrightarrow{F}\,.\,\overrightarrow{S}\,\,=\,\,F\times S\text{ }cos\,\theta \] \[=\,\, F\,\,\times \,\,S cos 9{{0}^{o}}= 0\] Example: (i) When an electron revolve around the nucleus in hydrogen atom in a particular orbit, it neither absorb nor emit any energy means its energy remains constant. (ii) When a satellite established once in a orbit around the earth and it starts revolving with particular speed, then no fuel is required for its circular motion. Sample problem based on work done Problem 121. A particle does uniform circular motion in a horizontal plane. The radius of the circle is 20 cm. The centripetal force acting on the particle is 10 N. It?s kinetic energy more...