A thin solid disk of mass m, initially rolling without slipping at speed √80 m s⁻¹, ascends a movable smooth wedge of mass m, initially stationary. The disk–wedge contact is frictionless. Find maximum vertical rise (m, g = 10 m s⁻²). [See Image.]
Physics · Mixed Chapter
JEE Main Practice

Question

A thin solid disk of mass m, initially rolling without slipping at speed √80 m s⁻¹, ascends a movable smooth wedge of mass m, initially stationary. The disk–wedge contact is frictionless. Find maximum vertical rise (m, g = 10 m s⁻²). [See Image.]

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Initial disk speed is v=√80 m s⁻¹. Because disk–wedge contact is frictionless, disk spin remains unchanged as it ascends. Step 2 — Horizontal momentum of disk+wedge is conserved. At highest point of rise, both masses m have same horizontal speed V. From mv=2mV, V=v/2. Step 3 — Initial translational energy (1/2)mv²; at maximum height common translational energy is (1/2)(2m)(v/2)²=mv²/4. Step 4 — Rotational kinetic energy cancels from initial and final states. Therefore mgH = mv²/2−mv²/4…