Fyneman exercise I ch10-1

Q.

Two gliders are free to move in a horizontal air through. One is stationary and the other collides with it perfectly elastically. They rebound with equal and opposite velocities. What is the ratio of their masses?

A.

Let moving glider’s mass $m_1$, velocity $v_{1}$ and stationary one’s mass $m_2$, velocity $v_{2}$.

Initially $v_{2i} = 0$ and total momentum is

$m_1 v_{1i}\tag{1}$

Kinetic energy is

$\frac{1}{2}m_1v_{1i}^2\tag{2}$

Finally, $-v_{1f} = v_{2f}$ and so total momentum is

$m_1v_{1f}+m_2 v_{2f}=m_1v_{1f}-m_2v_{1f}\tag{3}$

Kinetic energy is

$\frac{1}{2}m_1v_{1f}^2+\frac{1}{2}m_2v_{1f}^2\tag{4}$

Since elastic collision, momentum and kinetic energy are conserved, which means $(1)=(3)$ and $(2)=(4)$.

$m_1 v_{1i}=m_1 v_{1f}-m_2v_{1f}\tag{5}$

$\frac{1}{2}m_1 v_{1i}^2=\frac{1}{2}m_1 v_{1f}^2+\frac{1}{2}m_2 v_{1f}^2\tag{6}$

From eq.(5)

$v_{1i}=(1-\frac{m_2}{m_1}) v_{1f}\tag{7}$

From eq.(6)

$v_{1i}^2= (1+\frac{m_2}{m_1})v_{1f}^2\tag{8}$

Let $ratio=\frac{m_2}{m_1}=x$ and put $v_{1i}$ from eq.(7) to eq.(8) gives

$x^2-3x=0$

Therefore, ratio $x = 0$ or $3$. Since $m_2 \neq 0$,

ratio $x=3$