NEET Physics Two Dimensional Motion Momentum

Momentum

Category : NEET

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: (b) Change in momentum between complete projectile motion \[= 2mu sin\,\theta \] \[=2\times 0.5\times 98\times \sin 30{}^\circ \]= 49 N?s. Problem 15. A particle of mass 100 g is fired with a velocity 20 m sec?1 making an angle of 30o with the horizontal. When it rises to the highest point of its path then the change in its momentum is (a) \[\sqrt{3}kg\,m\,{{\sec }^{-1}}\]        (b) \[1/2 kg m se{{c}^{1}}\]      (c) \[\sqrt{2}\,kg\,m\,{{\sec }^{-1}}\]            (d) \[1\text{ }kg\text{ }m\text{ }se{{c}^{1}}\] Solution: (d) Horizontal momentum remains always constant So change in vertical momentum (D\[\vec{p}\]) = Final vertical momentum ? Initial vertical momentum \[=0-mu\sin \theta \] \[|\Delta P|\,=0.1\times 20\times \sin {{30}^{o}}\] \[=1\,kg\,m/sec\]. Problem 16. Two equal masses (m) are projected at the same angle (q) from two points separated by their range with equal velocities (v). The momentum at the point of their collision is (a) Zero             (b) \[2\text{ }mv\text{ }cos\,\theta \]                    (c) \[\,2\text{ }mv\text{ }cos\,\theta \]                  (d) None of these Solution: (a) Both masses will collide at the highest point of their trajectory with equal and opposite momentum. So net momentum of the system will be zero. Problem 17. A particle of mass m is projected with velocity v making an angle of \[{{45}^{o}}\] with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where g = acceleration due to gravity) [MP PMT 1994; UPSEAT 2000; MP PET 2001] (a) Zero             (b) \[m{{v}^{3}}/(4\sqrt{2}g)\]  (c) \[m{{v}^{3}}/(\sqrt{2}g)\]                (d) \[m{{v}^{2}}/2g\] Solution: (b) \[L=\frac{m\,\,{{u}^{3}}\cos \theta {{\sin }^{2}}\theta }{2g}=\frac{m{{v}^{3}}}{(4\sqrt{2}\,g)}\]       \[[As\,\theta = 4{{5}^{o}}]\] Problem 18. A body is projected from the ground with some angle to the horizontal. What happens to the angular momentum about the initial position in this motion [AIIMS 2000] (a) Decreases                                          (b) Increases (c) Remains same                                    (d) First increases and then decreases Solution: (b)       Problem 19. In case of a projectile, where is the angular momentum minimum (a) At the starting point (b) At the highest point (c) On return to the ground (d) At some location other than those mentioned above Solution: (a)


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