Engineering Mechanics Two balls of equal mass and of perfectly elastic material are lying on the floor. One of the balls with velocity ‘v’ is made to strike the second ball. Both the balls after impact will move with a velocity v/8 v/2 v/4 v v/8 v/2 v/4 v ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The loss of kinetic energy during inelastic impact, is given by (where m1 = Mass of the first body,m2 = Mass of the second body, and u1 and u2 = Velocities of the first and second bodies respectively.) [m₁ m₂/2(m₁ + m₂)] (u₁² - u₂²) [2(m₁ + m₂)/m₁ m₂] (u₁² - u₂²) [m₁ m₂/2(m₁ + m₂)] (u₁ - u₂)² [2(m₁ + m₂)/m₁ m₂] (u₁ - u₂)² [m₁ m₂/2(m₁ + m₂)] (u₁² - u₂²) [2(m₁ + m₂)/m₁ m₂] (u₁² - u₂²) [m₁ m₂/2(m₁ + m₂)] (u₁ - u₂)² [2(m₁ + m₂)/m₁ m₂] (u₁ - u₂)² ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The unit of moment of inertia of an area is kg-m2 kg/m2 kg-m-s2 m4 kg-m2 kg/m2 kg-m-s2 m4 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The moment of inertia of a thin disc of mass ‘m’ and radius ‘r’, about an axis through its center of gravity and perpendicular to the plane of the disc is mr²/6 mr²/2 mr²/8 mr²/4 mr²/6 mr²/2 mr²/8 mr²/4 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics If ‘P’ is the force acting on the body, ‘m’ is the mass of the body and ‘a’ is the acceleration of the body, then according to Newton's second law of motion, P + m.a = 0 P/m.a = 0 P - m.a = 0 P × m.a = 0 P + m.a = 0 P/m.a = 0 P - m.a = 0 P × m.a = 0 ANSWER DOWNLOAD EXAMIANS APP
Engineering Mechanics The wheels of a moving car possess Kinetic energy of translation only Kinetic energy of rotation only Potential energy only Kinetic energy of translation and rotation both Kinetic energy of translation only Kinetic energy of rotation only Potential energy only Kinetic energy of translation and rotation both ANSWER DOWNLOAD EXAMIANS APP
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