Worked example — collision via a spring

Problem

Ball 1 ($3\;\text{kg}$, $v = 6\;\text{m/s}$) approaches ball 2 ($5\;\text{kg}$, $v = -2\;\text{m/s}$). Between them there is a spring ($K = 200\;\text{N/m}$, $\ell_0 = 5\;\text{m}$, $\ell_A = \ell_0$). Find the maximum compression of the spring.

Solution

Velocity of the centre of mass (conserved throughout the process): $v_\text{CM} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2} = \frac{3\cdot 6 + 5\cdot(-2)}{8} = \frac{18 - 10}{8} = 1\;\text{m/s}$

State B (maximum compression). At that instant the two bodies move at the same velocity $v_\text{CM}$: the internal kinetic energy (of relative motion) is zero and has all been transferred to the spring.

Energy table (from A to B; $E$ in $\text{J}$):

	$E_\text{kin,CM}$	$E_\text{kin,int}$	$E_\text{pot,spring}$	$E_\text{tot}$
A	$\tfrac{1}{2}\cdot 8\cdot 1^2 = 4$	$\tfrac{1}{2}\cdot 3\cdot 5^2 + \tfrac{1}{2}\cdot 5\cdot 3^2 = 60$	$0$	$64$
B	$4$	$0$	$100(\ell_B - 5)^2$	$64$

(in A the velocities relative to the CM are $6-1 = 5$ and $-2-1 = -3$, giving $E_\text{kin,int} = 37{,}5 + 22{,}5 = 60\;\text{J}$; the spring energy is $\tfrac{1}{2}K x^2 = 100\,x^2$ with $x = \lvert\ell_B - 5\rvert$).

Maximum compression: $100(\ell_B - 5)^2 = 60 \;\Rightarrow\; \lvert\ell_B - 5\rvert = \sqrt{0{,}6} = 0{,}77\;\text{m}$ $\ev{5 - \ell_B \approx 0{,}77\;\mathrm{m}}$

From A to C (the spring returns to $\ell_0$ and gives back all its energy): the overall process is an elastic collision, so the trick $v_{1A} + v_{1C} = v_{2A} + v_{2C}$ applies, and together with conservation of $\vv{p}$ gives the final separation velocities $v_{1C}$ and $v_{2C}$.

Links

Topics: Quantità di moto e urti Concepts: Urto elastico · Conservazione della quantità di moto · Energia cinetica · Energia potenziale elastica · Centro di massa Skills: Metodo delle due tabelle Objects: Molla