Question

A simple pendulum of length 0.2 m has bob of mass 5 gm, it is pulled aside through an angle ${60}^0$ from the vertical. A spherical body of mass 2.5 gm is placed at the lowest position of the bob. When the bob is released it strikes the spherical body and comes to rest. What is the velocity of the spherical body?(g=9.8 ms-2) (in m/s)

• A. $1.4$
• B. $2.8$
• C. $3.5$
• D. $4.9$

Answer 1

The correct option is B $2.8$

Velocity at lowest point of pendulum before collision

$mgh=\frac{1}{2}m{v}^2$

$9.8\times 0.1=\frac{1}{2}\times v^2$

$v^2=1.96$
$v=1.4$ m/s. i.e.,
momentum conservation
$5\times 1.4=2.5\times v$
$\Rightarrow v=2.8$ m/s.