Question

A cylinder is rolling over frictionless horizontal surface with velocity v₀ as shown in figure. Coefficient of friction between wall and cylinder is m= $\frac{1}{4}$. If the collision between cylinder and wall is completely inelastic, then kinetic energy of cylinder after collision –

• A. Zero
• B. $\frac{mv_{0}^{2}}{32}$
• C. $\frac{mv_{0}^{2}}{4}$
• D. $\frac{3mv_{0}^{2}}{32}$

Answer 1

Answer: (D)

$\frac{3mv_{0}^{2}}{32}$

Answer 2

Collision between wall and cylinder is completely inelastic

\ $\int_{}^{}{Ndt}$= $mv_{0}$

$\int_{}^{}{fdt}$= mv_y [v_y : f cylinder in

upward direction after

Ž v_y = mv₀

Now, Angular impulse due to friction force

$\frac{mR^{2}\omega}{2}$–$\frac{mR^{2}}{2}$.$\frac{v_{0}}{R}$= $–\int_{}^{}{f.R.dt}$

[w : Angular velocity of cylinder after collision]

Ž w = $\frac{v_{0}(1 - 2\mu)}{R}$

Kinetic energy after collision

= $\frac{1}{2}mv^{2}$+$\frac{1}{2}I\omega^{2}$= $\frac{3}{32}mv_{0}^{2}$