Home › JEE Advanced › Physics › Rotational Motion › A solid sphere rolls at speed $v$ on a horizonta…
A solid sphere rolls at speed $v$ on a horizontal surface and elastically collides with an identical stationary sphere (line of centres horizontal). After friction restores rolling on both, their speeds are:
AFirst sphere $0$, second sphere $v$
BBoth $v/2$
CFirst sphere $2v/7$, second sphere $5v/7$ (momentum still $v$)
DFirst sphere $v$, second sphere $v$
Answer & Solution
Correct answer: C. First sphere $2v/7$, second sphere $5v/7$ (momentum still $v$)
Elastic identical collision transfers translation only: just after collision, sphere 1 has $v=0$ but still spins ($\omega = v/R$); sphere 2 has $v$ but no spin. Friction restores rolling: for sphere 2, friction decelerates $v$ and spins it up — final $v_2' = (I/(I+mR^2))\cdot 0 + (mR^2/(I+mR^2))\cdot v = (5/7)v$. For sphere 1, friction accelerates $v$ from $0$, final $v_1' = (2/7)v$ (mirror calculation). Momentum sum $(2/7)v + (5/7)v = v$ ✓.
Related questions
A bicycle wheel of radius $R$ rolls without slipping at linear speed $v$. The angular speeAn ice skater spinning with arms outstretched at $\omega_1$ pulls arms in, halving her momA solid disc of mass $M$, radius $R$, rotating about its centre axis has moment of inertiaA child pushes a door of width $0.8$ m perpendicular to the door at the handle with $20$ NMoment of inertia of an annular disc (inner radius r1, outer r2) about perpendicular axis For a solid cylinder rolling without slipping, the friction force at the contact point:A merry-go-round (I = 1000 kg m²) rotates at 0.5 rad/s. A 50 kg child runs and jumps onto A solid cylinder (I = MR²/2) of mass 2 kg, radius 0.1 m, rolls without slipping at 6 m/s.