USAPhO 1999, Четвъртфинал — Задача 3

• a. 0.05 m
• b. After the second collision m₂ stops and m₁ reaches the maximum height hᵢ = 0.20 m.

3. Let $h_i = 0.20\ \mathrm{m}$ the initial height of $m_1$.

$h_f$ = the final height of both masses.

$v_{1i}$ = the speed of $m_1$ just before the collision.

$v_{1f}$ = the speed of $m_1$ just after the collision.

$v_{2f}$ = the speed of $m_2$ just after the collision.

a. Before the collision, for $m_1$ $\quad m_1gh_i = \tfrac{1}{2}m_1v_{1i}^2$ [3]

After the collision, for $m_1$ $\quad m_1gh_f = \tfrac{1}{2}m_1v_{1f}^2$

After the collision, for $m_2$ $\quad m_2gh_f = \tfrac{1}{2}m_2v_{2f}^2$

Since both masses rise to the same height after the collision, they must have equal speeds after the collision $\quad v_{1f} = v_{2f} = v_f$ [3]

Since the collision is elastic, the total kinetic energy is unchanged [2]

$\tfrac{1}{2}m_1v_{1i}^2 = \tfrac{1}{2}m_1v_f^2 + \tfrac{1}{2}m_2v_f^2$ [2]

or $\quad m_1gh_i = m_1gh_f + m_2gh_f$

$m_1h_i = (m_1 + m_2)h_f \qquad \text{Eq. 1}$ [2]

Using momentum conservation [2]

$m_1v_{1i} = m_2v_f - m_1v_f$ [2]

or $\quad m_1\sqrt{2gh_i} = (m_2 - m_1)\sqrt{2gh_f}$

$m_1\sqrt{h_i} = (m_2 - m_1)\sqrt{h_f}$

Squaring both sides $\quad m_1^2h_i = \left(m_2^2 - 2m_1m_2 + m_1^2\right)h_f$

Substituting in Eq. 1 $\quad m_1(m_1 + m_2)h_f = \left(m_2^2 - 2m_1m_2 + m_1^2\right)h_f$

Solving for the relation between the masses $\quad 0 = m_2^2 - 3m_2m_1$

Or assuming $m_2 \neq 0$ $\quad m_2 = 3m_1$ [2]

Solving Eq. 1 for $h_f$ $\quad h_f = \dfrac{m_1}{m_1 + m_2}h_i = \dfrac{m_1}{m_1 + 3m_1}h_i = \dfrac{1}{4}0.20\ \mathrm{m} = 0.050\ \mathrm{m}$ [2]

b. After the second collision $m_2$ stops [3]

and $m_1$ reaches maximum height $h_i$ [2]