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Q3.12:
A ball is dropped from a height of $90 m$ on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between $t = 0$ to $12 s$.

Solution :

Step 1: Motion under Gravity (Initial Fall)

The ball is dropped from a height $h = 90 \, \text{m}$ with initial velocity $u = 0$. Using the kinematic equation $v^2 - u^2 = 2gh$ [Per Kinematic Law of Motion], where $g \approx 10 \, \text{m/s}^2$:

$v_1 = \sqrt{2 \times 10 \times 90} = \sqrt{1800} \approx 42.43 \, \text{m/s}$

The time taken for the first fall ($t_1$) is derived from $v = u + gt$:

$42.43 = 0 + 10 \times t_1 \implies t_1 = 4.24 \, \text{s}$

Step 2: Analysis of Subsequent Collisions

Upon collision, the ball loses $1/10$ of its speed. The new rebound velocity ($v_{up}$) is:

$v_{up} = v_{down} \times (1 - 0.1) = 0.9 \times v_{down}$

Since the motion is symmetric in a vacuum [Per Principle of Conservation of Energy], the time taken to rise to maximum height and return to the floor is $t_{rise} + t_{fall} = 2 \times (v_{up}/g)$.

Collision Velocity After Impact ($v_{up}$) Time for Round Trip ($2 \times v/g$) Cumulative Time (s)
0 $42.43 \, \text{m/s}$ $4.24 \, \text{s}$ (fall only) $4.24$
1 $38.19 \, \text{m/s}$ $7.64 \, \text{s}$ $11.88$

Step 3: Speed-Time Graph Construction

Time (s) Speed (m/s) 4.24 11.88

Step 4: Analytical Justification

  • Interval $0 < t < 4.24 \, \text{s}$: Speed increases linearly from $0$ to $42.43 \, \text{m/s}$ due to constant acceleration $g$.
  • Collision 1 at $t = 4.24 \, \text{s}$: Velocity instantaneously changes from $+42.43$ to $-38.19$ (downward magnitude).
  • Interval $4.24 < t < 11.88 \, \text{s}$: Velocity decreases from $38.19$ to $0$ (ascending), then increases to $38.19$ (descending) [Per Law of Conservation of Momentum for impulsive collisions].

Final Solution: The speed-time graph exhibits a series of sawtooth-like patterns where the peak speed decreases by a factor of $0.9$ at each interval of $t_n = 2(0.9^{n-1} \times 4.24)$, converging toward zero as $t$ increases.


More Questions from Class 11 Physics Motion in a Straight Line EXERCISES


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