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 .)

Question

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 .) (A) 10 ms- 1 (B) 20 ms- 1 (C) 40 ms- 1 (D) 50 ms- 1

Tardigrade Admin · Accepted Answer

With respect to elevator, Initial velocity of ball =5 m/s , downwards Let velocity of ball just before striking the elevator be v , then v2=52+2× 10× 100 =25+2000=2025 v=45 m/s Since the collision is elastic, Velocity of approach = velocity of separation ∴ Velocity of separation =45 m/s ∴ Velocity of ball after collision =50 m/s (w.r.t. ground)

Q. A heavy elevator is moving upwards with constant velocity $5 m se c^{- 1}$ . 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, $g = 10 m s^{- 2}$ .)

$10 m s^{- 1}$

$20 m s^{- 1}$

$40 m s^{- 1}$

$50 m s^{- 1}$

Q. A heavy elevator is moving upwards with constant velocity 5msec−1 . At time t=0 , a ball situated at a distance of 100m from the elevator is dropped. What will be the velocity of ball just after the collision? (Consider the collision to be elastic. Use, g=10ms−2 .)

10ms−1

20ms−1

40ms−1

50ms−1

Q. A heavy elevator is moving upwards with constant velocity $5 m se c^{- 1}$ . 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, $g = 10 m s^{- 2}$ .)

$10 m s^{- 1}$

$20 m s^{- 1}$

$40 m s^{- 1}$

$50 m s^{- 1}$