Question

A cylinder is rolling over frictionless horizontal surface with velocity $v_0$ as shown in figure. Coefficient of friction between wall and cylinder is $\mu =\frac{1}{4}.$ If the collision between cylinder and wall is completely inelastic, then kinetic energy of cylinder after collision.

• A. Zero
• B. $\frac{m{v}_0^2}{32}$
• C. $\frac{m{v}_0^2}{4}$
• D. $\frac{3m{v}_0^2}{32}$

Answer 1

The correct option is D $\frac{3m{v}_0^2}{32}$

Collision between wall and cylinder is completely inelastic.
$\therefore \int Ndt=m{v}_0$ and $\int fdt=m{v}_y$
$v_y$: velocity of cylinder in upward direction after collision]
$\Rightarrow v_y=\mu v_0$
Now, Angular impulse due to friction force
$\frac{m{R}^2\omega }{2}-\frac{m{R}^2}{2}.\frac{v_0}{R}=-\int f.R.dt$
$[\omega$: Angular velocity of cylinder after collision]
$\Rightarrow \omega =\frac{v_0(1-2\mu )}{R}\{\because \text{Given}\mu =\frac{1}{4}\}$
Kinetic energy after collision
$=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=\frac{3}{32}m{v}_0^2$