Fyneman exercise I ch10-2

Q.

Two equally massive gliders, moving in a level air trough at equal and opposite velocities, $v$ and $-v$, collide almost elastically, and rebound with slightly smaller speeds. They lose a fraction $f << 1$ of their kinetic energy in the collision. If these same gliders collide with one of them initially at rest, with what speed will the second glider move after the collision? (This small residual speed $\Delta v$ may easily be measured in terms of the final speed $v$ of the originally stationary glider, and thus the elasticity of the spring bumpers may be determined.)

NOTE : If $x<<1, \ \sqrt{1-x}\approx 1-\frac{1}{2}x$

A.

Let initial speed of glider-1 $v_i$, final speed of the other glider-2 initially at rest $v$, and final speed of glider-1 $\Delta v$.

Suppose $\Delta v$ is opposite $v$ .

Kinetic Energy

$\frac{1}{2}m(1-f)v_i^2=\frac{1}{2}m(v^2+\Delta v^2)$

$(1-f)v_i^2=v^2+\Delta v^2\tag{1}$

Momentum

$mv_i=m(v-\Delta v)$

$v_i=v-\Delta v\tag{2}$

From eq.(1) & eq.(2)

$(1-f)(v-\Delta v)^2=v^2+\Delta v^2$

$-2v\Delta v -f(v-\Delta v)^2=0\tag{3}$

Eq.(3) is not possible, because $v$, $\Delta v$, and $(v-\Delta v)^2$ are all positive.

We can conclude two glider moves same direction after collision.

Therefore eq.(2) should be

$v_i=v+\Delta v\tag{4}$

From eq.(1) & (4)

$2v\Delta v -f(v-\Delta v)^2 = 0$

$1-2(1+\frac{1}{f})\frac{\Delta v}{v}+(\frac{\Delta v}{v})^2 = 0$

$\frac{\Delta v}{v} = 1+\frac{1}{f}\pm \sqrt{(1+\frac{1}{f})^2-1}$

Since $\frac{1}{f} >> 1$ and $\frac{\Delta v}{v} \lt 1$

$\frac{\Delta v}{v}=1+\frac{1}{f}- \sqrt{(1+\frac{1}{f})^2-1}$

Let $1+\frac{1}{f} = y$

$\frac{\Delta v}{v}=y-y\sqrt{1-(\frac{1}{y})^2}\tag{5}$

$y >> 1(\because f<<1)$, so $\frac{1}{y} <<1$

Let $\frac{1}{y} = x << 1$, then eq.(5) becomes

$\frac{\Delta v}{v}=y-y\sqrt{1-x^2}\tag{6}$

$\sqrt{1-x^2}=\sqrt{1-x}\sqrt{1+x}\approx(1-\frac{1}{2}x)(1+\frac{1}{2}x)=1-\frac{1}{4}x^2$

Put above approximation to Eq.(6)

${\Delta v \over v} \approx y-y(1-{1 \over 4y^2})={1 \over 4y}$

${\Delta v \over v} \approx {1 \over 4}{f \over 1+f} \approx {1 \over 4}f$

$\Delta v = {vf \over 4}$