A uniform electric field of 15 N C⁻¹ makes an angle of 30° with the horizontal. A ball of charge 2 C, mass 3 kg and coefficient of restitution with the ground 0.5 is projected at 20 m s⁻¹ at 30° with the horizontal along the electric field. The horizontal distance travelled between the first and second collisions is 70√x m. Find x. (g = 10 m s⁻².)
Physics · Mixed Chapter
JEE Mains

Question

A uniform electric field of 15 N C⁻¹ makes an angle of 30° with the horizontal. A ball of charge 2 C, mass 3 kg and coefficient of restitution with the ground 0.5 is projected at 20 m s⁻¹ at 30° with the horizontal along the electric field. The horizontal distance travelled between the first and second collisions is 70√x m. Find x. (g = 10 m s⁻².)

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Explanation: Horizontal electric acceleration is aₓ=(qE/m)cos30°=5√3, and net vertical acceleration aᵧ=(qE/m)sin30°−g=−5 m/s². Initial velocity components are 10√3 and 10. First flight duration=4 s, so first-impact horizontal speed=30√3. Vertical impact velocity=−10; after restitution e=0.5 upward speed=5. Second-flight duration=2 s. Horizontal distance=30√3×2+(1/2)(5√3)×(2²)=70√3 m, so x=3.