Question

A rod of mass M and length L falls from rest from a height H making angle $\theta ={37}^{\circ }$ with horizontal as shown in figure. Coefficient of restitution between rod and ground is 0.5 then just after impact with ground.

• A. Angular velocity of rod is $\frac{\sqrt{2gH}}{L}$
• B. Speed of center C of rod is $\frac{111\sqrt{2gH}}{121}$
• C. Impulse on rod by ground is $\frac{50M\sqrt{2gH}}{121}$
• D. Direction of velocity of C just after impact is upwards.

Answer 1

Answer: (B), (C)

The correct options are
B Speed of center C of rod is $\frac{111\sqrt{2gH}}{121}$
C Impulse on rod by ground is $\frac{50M\sqrt{2gH}}{121}$


$V_0=\sqrt{2gH}$

By Impulse momentum relation,
$J=M(V_0-V)\dots \dots \left(1\right)$

Byrelation between angular impulse & angular momentum ,
$Jcos\theta \frac{L}{2}=\frac{M{L}^2}{12}\omega \dots \dots \left(2\right)$

By velocity of seperation and velocity of approach relation,
$0.5=\frac{\frac{L}{2}\omega cos\theta -v}{v_0}\dots \dots \left(3\right)$

Solving (1), (2) and (3) we get

$\omega =\frac{240\sqrt{2gH}}{|2|L}$, $J=\frac{50\sqrt{2gH}M}{121}$ and $V=\frac{111\sqrt{2gH}}{121}$