11th Physics Volume

( ) W J ( ) [ ] . A body of mass 10kg at rest is subjected to a force of 16N. Find the kinetic energy at the end of s. Solution: Mass m = kg Force F = N time t = s a = F m m s we know that, v = u + at = + .

× = m s – Kinetic energy K.E = mv J . A body of mass kg is thrown up vertically with a kinetic energy of J. If acceleration due to gravity is m s − , find the height at which the kinetic energy becomes half of the original value. Solution: Mass m = kg K.E E = J g = m s − At a height ‘h’, mgh = E × × h h .

Two bodies of masses kg and kg moving in the same direction along straight line with velocity cm s − and cm s − Appendix respectively suffer one dimensional elastic collision. Find their velocities after collision. Solution: Mass m = kg Mass m = kg V cms V cm s Solution: m m u Substituting the values, we get, v ( ) cms [ ] Likewise, cms × × × ( ) [ ] . A particle strikes a horizontal frictionless floor with a speed u at an angle θ with the vertical and rebounds with the speed v at an angle f with the vertical.

The coefficient of restitution between the particle and floor is e. What is the magnitude of v? θ φ Solution: Applying component of velocities, v sin φ u cos θ u sin θ v cos φ θ θ φ The x - component of velocity is usin q = vsin f ( ) The magnitude of y – component of velocity is not same, therefore, using coefficient of restitution, e cos cos ( ) Squaring ( ) and ( ) and adding we get e u adding e u sin sin cos cos sin cos e e sin cos sin cos Appendix . A particle of mass m is fixed to one end of a light spring of force constant k and un-stretched length l .

It is rotated with an angular velocity w in horizontal circle. What will be the length increase in the spring? Solution: Mass of particle = m Force constant = k Un-stretched length = l Angular velocity = w F s x = l Let ‘x’ be the increase in the length of the spring. The new length = ( l +x) = r When the spring is rotated in a horizontal circle, Spring force = centripetal force.

kx = m w ( l +x) x m l k–m . A gun fires bullets per second into a target X. If the mass of each bullet is g and its speed ms − , then calculate the power delivered by the bullets. Solution: Power = work done per second = total kinetic energy of bullets per second P (kinetic energy of each bullet per second) P = W P = .

kW SOLVED EXAMPLE UNIT- . Three particles of masses m = kg, m = kg and m = kg are placed at the corners of an equilateral triangle of side 1m as shown in Figure. Find the position of centre of mass. Solution: m m m ( , ) ( √ ) G C D , The centre of mass of an equilateral triangle lies at its geometrical centre G.

The positions of the mass m , m and m are at positions A, B and C as shown in the Figure. From the given position of the masses, the coordinates of the masses m and m are easily marked as ( , ) and ( , ) respectively. Appendix To find the position of m the Pythagoras theorem is applied. As the ∆DBC is a right angle triangle, BC CD DB CD BC DB CD CD = The position of mass m is , or .

, . X Coordinate of centre of mass, m x m x m x CM x CM ( ) ( ) ( . ) x CM = m Y Coordinate of centre of mass, m y m y m y CM y CM ( ) ( ) ( ) y CM = m. ∴ The coordinates of centre of mass G CM CM ,