Question

Two balls of masses $m_1$ and $m_2$ are placed on top of one over the other (with a small gap between them) and then dropped on to the ground. What is the ratio $\frac{m_1}{m_2}$ for which the upper ball ultimately receives the largest possible fraction of the total energy? Take all collisions as elastic. Neglect air resistance.

• A. 1:1
• B. 1:2
• C. 1:3
• D. 1:4

Answer 1

Answer: (C)

1:3

Answer 2

To solve this problem, we need to analyze the collisions involved. There are two main collisions:

1. The lower ball ($m_2$) collides with the ground.
2. The upper ball ($m_1$) collides with the lower ball ($m_2$).

Let's assume the balls are dropped from a height $h$. Just before hitting the ground, both balls will have a downward velocity $v_0$, where $v_0 = \sqrt{2gh}$.

Step 1: Collision of the lower ball ($m_2$) with the ground.

The collision with the ground is elastic. This means the ball bounces back with the same speed but in the opposite direction. So, just after hitting the ground, the lower ball $m_2$ moves upwards with velocity $v_0$.

Step 2: Collision between the upper ball ($m_1$) and the lower ball ($m_2$).

Immediately after $m_2$ bounces off the ground, it moves upwards with velocity $v_0$. At this instant, the upper ball $m_1$ is still moving downwards with velocity $v_0$ (since there's a small gap, it hasn't hit $m_2$ yet).

Let's define the upward direction as positive.

Before the collision between $m_1$ and $m_2$:

• Initial velocity of $m_1$, $u_1 = -v_0$ (downwards)
• Initial velocity of $m_2$, $u_2 = v_0$ (upwards)

Since the collision is elastic, we can use the conservation of momentum and the coefficient of restitution ($e=1$) equations. Let the final velocities after the collision be $v_1$ for $m_1$ and $v_2$ for $m_2$.

1. Conservation of Momentum: $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ $m_1 (-v_0) + m_2 (v_0) = m_1 v_1 + m_2 v_2$ $(m_2 - m_1) v_0 = m_1 v_1 + m_2 v_2 \quad (Eq. 1)$

2. Coefficient of Restitution ($e=1$): $v_2 - v_1 = -(u_2 - u_1)$ $v_2 - v_1 = -(v_0 - (-v_0))$ $v_2 - v_1 = -2v_0$ $v_1 = v_2 + 2v_0 \quad (Eq. 2)$

Substitute $v_1$ from (Eq. 2) into (Eq. 1): $(m_2 - m_1) v_0 = m_1 (v_2 + 2v_0) + m_2 v_2$ $(m_2 - m_1) v_0 = m_1 v_2 + 2m_1 v_0 + m_2 v_2$ $(m_2 - m_1 - 2m_1) v_0 = (m_1 + m_2) v_2$ $(m_2 - 3m_1) v_0 = (m_1 + m_2) v_2$ $v_2 = \frac{m_2 - 3m_1}{m_1 + m_2} v_0$

Now, substitute $v_2$ back into (Eq. 2) to find $v_1$: $v_1 = \frac{m_2 - 3m_1}{m_1 + m_2} v_0 + 2v_0$ $v_1 = \left( \frac{m_2 - 3m_1 + 2(m_1 + m_2)}{m_1 + m_2} \right) v_0$ $v_1 = \left( \frac{m_2 - 3m_1 + 2m_1 + 2m_2}{m_1 + m_2} \right) v_0$ $v_1 = \left( \frac{3m_2 - m_1}{m_1 + m_2} \right) v_0$

Step 3: Maximize the fraction of total energy for the upper ball.

The initial total energy of the system (potential energy before dropping) is $E_{total} = (m_1+m_2)gh$. Since $v_0 = \sqrt{2gh}$, we have $gh = v_0^2/2$. So, $E_{total} = \frac{1}{2}(m_1+m_2)v_0^2$.

The final kinetic energy of the upper ball ($m_1$) after the collision is $KE_1 = \frac{1}{2}m_1 v_1^2$. We want to maximize the fraction $F = \frac{KE_1}{E_{total}}$. $F = \frac{\frac{1}{2}m_1 v_1^2}{\frac{1}{2}(m_1+m_2)v_0^2} = \frac{m_1 v_1^2}{(m_1+m_2)v_0^2}$

Let $x = \frac{m_1}{m_2}$. Then $m_1 = xm_2$. Substitute $m_1 = xm_2$ into the expression for $v_1$: $v_1 = \left( \frac{3m_2 - xm_2}{xm_2 + m_2} \right) v_0 = \left( \frac{m_2(3 - x)}{m_2(x + 1)} \right) v_0 = \left( \frac{3 - x}{1 + x} \right) v_0$

Now substitute this into the fraction $F$: $F(x) = \frac{xm_2 \left( \left( \frac{3 - x}{1 + x} \right) v_0 \right)^2}{(xm_2 + m_2) v_0^2} = \frac{x m_2 \left( \frac{3 - x}{1 + x} \right)^2 v_0^2}{m_2(x + 1) v_0^2}$ $F(x) = \frac{x}{(x+1)} \left( \frac{3 - x}{1 + x} \right)^2 = \frac{x(3-x)^2}{(1+x)^3}$

To find the maximum value of $F(x)$, we need to differentiate $F(x)$ with respect to $x$ and set the derivative to zero. $F(x) = \frac{x(9 - 6x + x^2)}{(1+x)^3} = \frac{9x - 6x^2 + x^3}{(1+x)^3}$ Using the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$: $u = 9x - 6x^2 + x^3 \implies u' = 9 - 12x + 3x^2$ $v = (1+x)^3 \implies v' = 3(1+x)^2$

$F'(x) = \frac{(9 - 12x + 3x^2)(1+x)^3 - (9x - 6x^2 + x^3)3(1+x)^2}{((1+x)^3)^2}$ $F'(x) = \frac{(1+x)^2 \left[ (9 - 12x + 3x^2)(1+x) - 3(9x - 6x^2 + x^3) \right]}{(1+x)^6}$ $F'(x) = \frac{(9 - 12x + 3x^2)(1+x) - 3(9x - 6x^2 + x^3)}{(1+x)^4}$

Set the numerator to zero to find critical points: $(9 - 12x + 3x^2)(1+x) - 3(9x - 6x^2 + x^3) = 0$ Expand the first term: $9 + 9x - 12x - 12x^2 + 3x^2 + 3x^3 = 9 - 3x - 9x^2 + 3x^3$ So, the equation becomes: $(9 - 3x - 9x^2 + 3x^3) - (27x - 18x^2 + 3x^3) = 0$ $9 - 3x - 9x^2 + 3x^3 - 27x + 18x^2 - 3x^3 = 0$ Combine like terms: $9 + (-3 - 27)x + (-9 + 18)x^2 + (3 - 3)x^3 = 0$ $9 - 30x + 9x^2 = 0$ Divide by 3: $3x^2 - 10x + 3 = 0$

Solve this quadratic equation for $x$: $x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(3)(3)}}{2(3)}$ $x = \frac{10 \pm \sqrt{100 - 36}}{6}$ $x = \frac{10 \pm \sqrt{64}}{6}$ $x = \frac{10 \pm 8}{6}$

Two possible values for $x$: $x_1 = \frac{10 + 8}{6} = \frac{18}{6} = 3$ $x_2 = \frac{10 - 8}{6} = \frac{2}{6} = \frac{1}{3}$

Let's check these values:

• If $x = 3$ (i.e., $m_1 = 3m_2$): $v_1 = \left( \frac{3 - 3}{1 + 3} \right) v_0 = 0$. If $v_1=0$, then $KE_1=0$, which means the upper ball receives no energy. This is a minimum.

• If $x = 1/3$ (i.e., $m_1 = m_2/3$): $v_1 = \left( \frac{3 - 1/3}{1 + 1/3} \right) v_0 = \left( \frac{8/3}{4/3} \right) v_0 = 2v_0$. Let's calculate the fraction of energy for $x=1/3$: $F(1/3) = \frac{1/3}{(1/3+1)} \left( \frac{3-1/3}{1+1/3} \right)^2 = \frac{1/3}{4/3} \left( \frac{8/3}{4/3} \right)^2$ $F(1/3) = \frac{1}{4} (2)^2 = \frac{1}{4} \cdot 4 = 1$. This means the upper ball receives 100% of the total energy, which is the largest possible fraction.

Thus, the ratio $\frac{m_1}{m_2}$ for which the upper ball receives the largest possible fraction of the total energy is $1/3$.