If the collision occurs at time t 0 = 2 , then the value of 4 b 2 a 2 will be _______. [2024]

Question

Two particles, 1 and 2, each of mass $m$ , are connected by a massless spring, and are on a horizontal frictionless plane, as shown in the figure. Initially, the two particles, with their center of mass at $x_{0}$ , are oscillating with amplitude $a$ and angular frequency $ω$ . Thus, their positions at time $t$ are given by $x_{1} (t) = (x_{0} + d) + a \sin ω t$ and $x_{2} (t) = (x_{0} - d) - a \sin ω t,$ respectively, where $d > 2 a .$ Particle 3 of mass $m$ moves towards this system with speed $u_{0} = \frac{a ω}{2},$ and undergoes instantaneous elastic collision with particle 2, at time $t_{0}$ . Finally, particles 1 and 2 acquire a center of mass speed $v_{cm}$ and oscillate with amplitude $b$ and the same angular frequency $ω$ .

Q. If the collision occurs at time $t_{0} = \frac{π}{2 ω},$ then the value of $\frac{4 b^{2}}{a^{2}}$ will be _______. [2024]

Smart Achievers · Accepted Answer

(4.25)

$If the collision occurs at time$ $t_{0} = \frac{π}{2 ω} = \frac{T}{4}$

Particles are at extreme position.

Using work-energy theorem, $W_{spring} = Δ K$

$\frac{1}{2} k {(2 b)}^{2} - \frac{1}{2} k {(2 a)}^{2} = 2 \times \frac{1}{2} m {(\frac{a ω}{4})}^{2}$

$4 k b^{2} - 4 k a^{2} = 2 \times m \times \frac{a^{2}}{16} \times \frac{2 k}{m}$

$\Rightarrow 4 b^{2} = \frac{17}{4} a^{2}$

$∴$ $\frac{4 b^{2}}{a^{2}} = 4.25$

Solution

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Solution

Ans. (4.25) If the collision occurs at time t0=π2ω=T4 Particles are at extreme position. Using work-energy theorem, Wspring=ΔK 12k(2b)2-12k(2a)2=2×12m(aω4)2 4kb2-4ka2=2×m×a216×2km ⇒4b2=174a2 ∴ 4b2a2=4.25

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