A sphere of mass m is moving with a velocity 3î + 4ĵ + 5k̂, hits a smooth wall and rebounds with velocity 2î + 3ĵ + 2k̂. Then the coefficient of restitution between the sphere and the wall is n/6. Find n = ?
Physics · Mixed Chapter
JEE Mains

Question

A sphere of mass m is moving with a velocity 3î + 4ĵ + 5k̂, hits a smooth wall and rebounds with velocity 2î + 3ĵ + 2k̂. Then the coefficient of restitution between the sphere and the wall is n/6. Find n = ?

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: For a smooth fixed wall the collision impulse is normal to its surface. The velocity difference defines that normal: Δv=v′−v=(−1,−1,−3). Step 2: Project incident and outgoing velocities on the (un-normalized) wall normal n=(1,1,3): v·n=3+4+15=22; v′·n=2+3+6=11. The outgoing normal direction is reversed; the coefficient of restitution is the ratio of the normal speed magnitudes, e=11/22=1/2. Step 3: e=n/6 in the question, so n=6×(1/2)=3.