A ball strikes a horizontal floor at an angle θ = 45° with the normal to the floor. The coefficient of restitution between the floor and the ball during the collision is e = 1/2. Fraction of kinetic energy of the ball lost in this collision is n/(n+5). Then n = ...
Physics · Mixed Chapter
JEE Mains

Question

A ball strikes a horizontal floor at an angle θ = 45° with the normal to the floor. The coefficient of restitution between the floor and the ball during the collision is e = 1/2. Fraction of kinetic energy of the ball lost in this collision is n/(n+5). Then n = ...

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: For a smooth horizontal floor, the tangential speed component is unchanged in impact and the normal component is reversed with magnitude multiplied by e=1/2. Step 2: Fractional kinetic-energy loss is (1−e²)cos²θ=(1−1/4)(1/2)=3/8 because θ=45° is measured from the floor normal. Step 3: Set n/(n+5)=3/8; then 8n=3n+15, giving n=3.