JEE MainPhysicsMotion in One Dimension
A ball is dropped from rest from a height of 20 m . It hits the ground and rebounds, losing half of its impact speed during the collision. Taking the upward direction as positive and g = 10 m s ⁻² , which of the following sequences of (t, v) coordinates correctly represents the key events on the velocity-time graph of the ball from t = 0 until it hits the ground the second time? (Assume the collision time is negligib
Options
- A(0, 0) (2, -20) , instantaneous jump to (2, 20) (4, 0) (6, -20)
- B(0, 0) (2, -20) , instantaneous jump to (2, 10) (3, 0) (4, -10)
- C(0, 0) (2, -20) , instantaneous jump to (2, -10) (3, -20) (4, -30)
- D(0, 0) (2, 20) , instantaneous jump to (2, 10) (3, 0) (4, 10)
Correct answer
B. (0, 0) (2, -20) , instantaneous jump to (2, 10) (3, 0) (4, -10)
Step-by-step solution
For the downward journey, the initial velocity is u = 0 and acceleration is a = -10 m s ⁻² . Time taken to fall 20 m is given by s = ut + 1 2 at² -20 = 0 - 1 2 (10)t² t = 2 s Velocity just before the first impact is v = u + at = 0 - 10(2) = -20 m s ⁻¹ . During the collision, the ball loses half its speed and reverses direction. The velocity just after impact is v' = +10 m s ⁻¹ . The ball then moves upwards with a = -10 m s ⁻² . The time to reach the highest point is t' = v' |a| = 10 10 = 1 s . So, at t = 2 + 1 = 3