Spring-mass systems — collision problems — Hinglish notes

1.3.13 · Physics › Work, Energy & Power

WHY split the problem into phases?

Phase 2 setup: "common velocity" insight

Common velocity derive karna (scratch se)

Mass $m_{1}$ ka block $u_{1}$ speed se move karta hua $m_{2}$ block (initially $u_{2}$ ) se hit karta hai jisme spring attached hai. Rightward positive lo.

Maximum compression par, dono $v$ se move karte hain. Momentum conserved hai (Phase 2 mein koi external horizontal force nahi):

$m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v$

Maximum compression $x_{ma x}$ derive karna

Energy before compression = energy at max compression (KE partly spring PE mein convert hoti hai):

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} (m_{1} + m_{2}) v^{2} + \frac{1}{2} k x_{ma x}^{2}$

Yeh step kyun? Total KE poori store nahi hoti — system phir bhi $v$ par as a whole move kar raha hai. Sirf centre-of-mass frame mein KE hi spring mein store ho sakti hai.

$x_{ma x}$ solve karo. $v$ substitute karke aur simplify karne par clean reduced-mass form milti hai:

$\frac{1}{2} k x_{ma x}^{2} = \frac{1}{2} μ (u_{1} - u_{2})^{2}, μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}$

Figure — Spring-mass systems — collision problems

Worked Example 1 — block hits a spring fixed to a wall

Ek $2 kg$ ka block $3 m/s$ se slide karta hua ek spring ( $k = 600 N/m$ ) se takraata hai jo wall se attached hai. Max compression dhundho.

Yahan momentum kyun nahi? Wall ek external force exert karta hai, isliye momentum conserved nahi hai. Lekin energy hai (frictionless, ideal spring).

$\frac{1}{2} m u^{2} = \frac{1}{2} k x^{2} \Rightarrow x = u m / k$ Yeh step kyun? Saari KE spring PE mein convert ho jaati hai kyunki wall block ko poori tarah rest mein le aata hai.

$x = 3 2/600 = 3 \times 0.0577 = 0.173 m$

Worked Example 2 — two free blocks, spring on one

$m_{1} = 2 kg$ at $u_{1} = 6 m/s$ hits $m_{2} = 4 kg$ (at rest) jo ek spring $k = 1200 N/m$ carry kar raha hai. (a) max compression par velocity, (b) max compression, (c) final velocities dhundho.

(a) Common velocity — Kyun? Max compression ⇔ same velocity. $v = \frac{2 ( 6 ) + 4 ( 0 )}{6} = 2 m/s$

(b) Max compression — Reduced mass kyun? Sirf relative-motion KE store hoti hai. $μ = \frac{2 \times 4}{6} = \frac{4}{3}, x = (6 - 0) \frac{4/3}{1200} = 6 \times 0.0333 = 0.200 m$

(c) Final velocities — Spring eventually poori tarah wapas spring back karta hai → encounter effectively ek perfectly elastic collision hai (spring saari energy restore kar deta hai). Elastic-collision results use karo: $v_{1}^{'} = \frac{m _{1} - m _{2}}{m _{1} + m _{2}} u_{1} = \frac{- 2}{6} (6) = - 2 m/s, v_{2}^{'} = \frac{2 m _{1}}{m _{1} + m _{2}} u_{1} = \frac{4}{6} (6) = 4 m/s$ Elastic kyun? Ek ideal spring energy store karta hai aur phir 100% wapas deta hai — koi loss nahi → elastic.

Worked Example 3 — spring stays partly compressed?

Agar blocks max compression par stuck/latched ho jaate hain (ek clip spring ko lock kar deta hai), toh collision ab perfectly inelastic hai: woh saath mein $v = v_{c m}$ par hamesha ke liye move karte hain, aur energy $\frac{1}{2} μ (u_{1} - u_{2})^{2}$ stored hai (return nahi hoti). Same $x_{ma x}$ , lekin final state alag hai.

Yeh kyun matter karta hai: "max compression" ka maths identical hai; jo cheez change hoti hai woh yeh hai ki stored energy release hogi (elastic outcome) ya trapped rahegi (inelastic outcome).

Common mistake Steel-manning the classic errors

Wrong idea 1: "Collision ke dauran bhi energy conservation use karo." Kyun sahi lagta hai: spring ideal hai, toh surely energy hamesha conserved hai. Fix: Instantaneous contact ke dauran, finite spring impulse ≈ 0, isliye momentum use karo. Energy tabhi safe hai use karne ke liye jab tum stored PE ko correctly track karo — aur sirf tab agar koi latch/friction use remove na kare.

Wrong idea 2: "Saari KE spring mein jaati hai, toh do free blocks ke liye $\frac{1}{2} m_{1} u_{1}^{2} = \frac{1}{2} k x^{2}$ ." Kyun sahi lagta hai: wall case mein kaam kiya. Fix: do free blocks ke saath system phir bhi $v_{c m}$ par move karta hai, isliye sirf CM-frame KE $= \frac{1}{2} μ (u_{1} - u_{2})^{2}$ store hoti hai. $\frac{1}{2} (m_{1} + m_{2}) v^{2}$ subtract karo.

Wrong idea 3: "Max compression tab hoti hai jab front block ruk jaata hai." Kyun sahi lagta hai: compression tab sabse badi hogi jab target sabse zyada move kare. Fix: compression tab tak badhti hai jab tak woh approach karte hain. Yeh tab maximum hoti hai jab unki velocities equal hoti hain, na ki jab koi ek zero ho.

Recall Feynman: explain to a 12-year-old

Socho ice par do carts hain, ek springy bumper ke saath. Jab woh bump karte hain, spring squish ho jaata hai. Yeh tab tak squish hota rehta hai jab tak ek cart doosri ko catch up kar rahi hoti hai. Squish tab sabse bada hota hai jab exactly woh dono same speed par roll kar rahe hote hain — phir spring unhe wapas alag push kar deta hai. Quick bump ke dauran, pushing itni fast hoti hai ki hum bas kehte hain "total push-power (momentum) same rehti hai." Ek perfect spring ek perfect trampoline hai: yeh har ek bit energy wapas deta hai, isliye carts bina koi energy khoye bounce off ho jaate hain.

Flashcards

Instantaneous collision ke dauran kaun sa conservation law apply hota hai, aur kyun?

Momentum — spring ki finite force tiny collision time mein negligible impulse deti hai.

Spring compression maximum kab hoti hai?

Jab dono blocks ki same velocity ho (relative velocity = 0).

m_{1} u_{1} + m_{2} u_{2}

ke liye max compression par common velocity kya hai?

v = \frac{m _{1} u _{1} + m _{2} u _{2}}{m _{1} + m _{2}}

(centre-of-mass velocity).

Do free blocks ke liye max compression ka formula?

x_{ma x} = (u_{1} - u_{2}) μ / k

jahan

μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}

x_{ma x}

mein reduced mass

μ

kyun aata hai?

Sirf relative-motion (CM-frame) KE

\frac{1}{2} μ (u_{1} - u_{2})^{2}

store ho sakti hai; relative motion ki effective mass

μ

hoti hai.

Block wall-fixed spring se takraata hai: kaun sa law aur result?

Sirf energy (wall = external force);

x = u m / k

Do free blocks ke saath ideal spring ke liye, overall outcome kis type ki collision hai?

Perfectly elastic — spring stored energy ka 100% wapas karta hai.

Agar ek latch spring ko max compression par lock kar de, toh yeh kis type ki collision hai?

Perfectly inelastic — energy

\frac{1}{2} μ (u_{1} - u_{2})^{2}

trapped rehti hai; blocks saath move karte hain.

Max compression par spring mein kitni KE store hoti hai?

\frac{1}{2} μ (u_{1} - u_{2})^{2}

(CM-frame kinetic energy).

Spring-mass systems — collision problems