If the collision occurs at time t0=0 the value of vcm/aw 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 $ω$ . [2024]

Q. If the collision occurs at time $t_{0} = 0,$ the value of $\frac{v_{cm}}{(a ω)}$ will be ________. [2024]

Smart Achievers · Accepted Answer

(0.75)

$At time t_{0} = 0 collision occurs$

$t$

$V_{C M} = \frac{m (\frac{a ω}{2}) + m (a ω)}{m + m} = \frac{3 a ω}{4}$

$∴$ $\frac{V_{C M}}{a ω} = \frac{3}{4} = 0.75$

Solution

Ans.
(0.75)

$At time t_{0} = 0 collision occurs$

$Before collision$

$V_{C M} = \frac{m (\frac{a ω}{2}) + m (a ω)}{m + m} = \frac{3 a ω}{4}$

$∴$ $\frac{V_{C M}}{a ω} = \frac{3}{4} = 0.75$

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Solution

Ans. (0.75) At time t0=0 collision occurs Before collision VCM=m(aω2)+m(aω)m+m=3aω4 ∴ VCMaω=34=0.75

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Ans.
(0.75)

$At time t_{0} = 0 collision occurs$

$Before collision$

$V_{C M} = \frac{m (\frac{a ω}{2}) + m (a ω)}{m + m} = \frac{3 a ω}{4}$

$∴$ $\frac{V_{C M}}{a ω} = \frac{3}{4} = 0.75$