Q. A heavy elevator is moving upwards with constant velocity $5 \, msec^{- 1}$ . At time $t=0$ , a ball situated at a distance of $100 \, m$ from the elevator is dropped. What will be the velocity of ball just after the collision? (Consider the collision to be elastic. Use, $g=10 \, ms^{- 2}$ .)
NTA AbhyasNTA Abhyas 2020
Solution:
With respect to elevator,
Initial velocity of ball $=5 \, m/s$ , downwards
Let velocity of ball just before striking the elevator be $v$ ,
then $v^{2}=5^{2}+2\times 10\times 100$
$=25+2000=2025$
$v=45 \, m/s$
Since the collision is elastic,
Velocity of approach $=$ velocity of separation
$\therefore $ Velocity of separation $=45 \, m/s$
$\therefore $ Velocity of ball after collision $=50 \, m/s$ (w.r.t. ground)
Initial velocity of ball $=5 \, m/s$ , downwards
Let velocity of ball just before striking the elevator be $v$ ,
then $v^{2}=5^{2}+2\times 10\times 100$
$=25+2000=2025$
$v=45 \, m/s$
Since the collision is elastic,
Velocity of approach $=$ velocity of separation
$\therefore $ Velocity of separation $=45 \, m/s$
$\therefore $ Velocity of ball after collision $=50 \, m/s$ (w.r.t. ground)