Question

A ball of mass 0.2 kg is thrown vertically down from a height of 10 m. It collides with the floor and loses 50% of its energy and then rises back to the same height. The value of its initial velocity is

• A. 14 ms^-1
• B. 196 ms^-1
• C. 20 ms^-1
• D. Zero

Answer 1

The Correct Option is A

Given: 

• Mass of the ball, \( m = 0.2 \) kg
• Initial height, \( h = 10 \) m
• Energy lost after collision = 50%
• Acceleration due to gravity, \( g = 9.8 \) m/s²

Step 1: Use Energy Conservation

Let initial velocity be \( u \). The total mechanical energy before hitting the ground:

\[ E_i = \frac{1}{2} m u^2 + mgh \]

After hitting the ground, the ball loses 50% of its energy and rises back to the same height, meaning its remaining energy is:

\[ E_f = \frac{1}{2} E_i \]

Step 2: Express in Terms of Initial Velocity

Since the ball reaches the same height after losing 50% energy, the energy left is equal to its potential energy at height \( h \):

\[ mgh = \frac{1}{2} \left( \frac{1}{2} m u^2 + mgh \right) \]

Solving for \( u \):

\[ 2mgh = \frac{1}{2} m u^2 + mgh \]

\[ mgh = \frac{1}{2} m u^2 \]

\[ u^2 = 2gh \]

\[ u = \sqrt{2 \times 9.8 \times 10} \]

\[ u = \sqrt{196} = 14 \text{ m/s} \]

Answer: The initial velocity of the ball is 14 m/s (Option A).

Answer 2

Let:

• m = mass of the ball = 0.2 kg
• h = initial height = 10 m
• u = initial downward velocity (what we need to find)
• v_before = velocity just before hitting the floor
• v_after = velocity just after hitting the floor
• h_final = final height reached after bouncing = 10 m
• g = acceleration due to gravity ≈ 9.8 m/s²

We can analyze the motion in three parts:

1. Upward motion after collision:

The ball rises to a height h_final = 10 m after bouncing. At the maximum height, its velocity is 0. We can use the conservation of energy (or kinematics) for this upward journey. The kinetic energy just after the bounce is converted into potential energy at the peak.

K.E_after = P.E_top

(1/2) * m * v_after² = m * g * h_final

Cancel out 'm':

(1/2) * v_after² = g * h_final

v_after² = 2 * g * h_final

v_after² = 2 * 9.8 m/s² * 10 m

v_after² = 196 m²/s²

v_after = sqrt(196) = 14 m/s

So, the velocity immediately after bouncing is 14 m/s upwards.

2. Collision with the floor:

The ball loses 50% of its energy during the collision. This energy is kinetic energy just before and just after the collision.

K.E_after = (1 - 0.50) * K.E_before

K.E_after = 0.5 * K.E_before

(1/2) * m * v_after² = 0.5 * (1/2) * m * v_before²

Cancel out (1/2) * m:

v_after² = 0.5 * v_before²

We know v_after² = 196 m²/s²:

196 = 0.5 * v_before²

v_before² = 196 / 0.5 = 196 * 2 = 392 m²/s²

So, the square of the velocity just before hitting the floor is 392 m²/s².

3. Downward motion before collision:

The ball is thrown downwards with initial velocity 'u' from height 'h'. We can use conservation of energy for this part. The total initial energy (potential + kinetic) equals the kinetic energy just before impact (potential energy is zero at the floor).

P.E_initial + K.E_initial = K.E_before

(m * g * h) + (1/2 * m * u²) = (1/2 * m * v_before²)

Cancel out 'm':

g * h + (1/2) * u² = (1/2) * v_before²

Substitute the known values:

(9.8 m/s² * 10 m) + (1/2) * u² = (1/2) * (392 m²/s²)

98 + 0.5 * u² = 196

0.5 * u² = 196 - 98

0.5 * u² = 98

u² = 98 / 0.5 = 98 * 2 = 196 m²/s²

u = sqrt(196) = 14 m/s

The initial velocity with which the ball was thrown downwards is 14 m/s.

Answer: The correct option is (A) 14 ms^-1.