Consider a horizontal surface moving vertically upward with velocity 2 m/s. A very small block of mass 2 kg is moving with velocity 2î − 2ĵ. If coefficient of restitution is 1/2 and friction coefficient is 1/3 then find horizontal component of velocity Vₓ of the block (in m/s) after collision.
Physics · Mixed Chapter
JEE Mains

Question

Consider a horizontal surface moving vertically upward with velocity 2 m/s. A very small block of mass 2 kg is moving with velocity 2î − 2ĵ. If coefficient of restitution is 1/2 and friction coefficient is 1/3 then find horizontal component of velocity Vₓ of the block (in m/s) after collision.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: The surface moves upward with velocity 2ĵ and the block initially has velocity 2î−2ĵ. Its incident vertical speed relative to the surface is −2−2=−4 m/s. Step 2: Coefficient of restitution e=1/2 makes the outgoing relative normal speed +2 m/s, so final vertical lab speed is vᵧ′=2+2=4 m/s. Step 3: The vertical velocity change is 4−(−2)=6 m/s. Maximum frictional reduction of the tangential velocity is μΔvₙ=(1/3)(6)=2 m/s. Step 4: Initial horizontal velocity is +2 m/s, exactly cancelled by…