A ball is moving at an angle 60° to the vertical when it collides with a smooth horizontal plane, perfectly inelastically. Its kinetic energy just before collision is 12 J. Find kinetic energy after impact (J).
Physics · Mixed Chapter
JEE Main Practice

Question

A ball is moving at an angle 60° to the vertical when it collides with a smooth horizontal plane, perfectly inelastically. Its kinetic energy just before collision is 12 J. Find kinetic energy after impact (J).

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Smooth collision preserves horizontal velocity component, while perfectly inelastic normal impact sets vertical component to zero. Step 2: Angle with vertical is 60°, so horizontal speed before collision is v sin60°. Step 3: Ratio of kinetic energies K_after/K_before =sin²60°=3/4. Step 4: K_after=(3/4)×12=9 J.