Two blocks each of the mass 𝑚 areconnected to a spring of spring constant𝑘. If both are given velocity 𝜐 in oppositedirections, then the maximumelongation of the spring is
Question
Two blocks each of the mass 𝑚 areconnected to a spring of spring constant𝑘. If both are given velocity 𝜐 in oppositedirections, then the maximumelongation of the spring is
Solution
The maximum elongation of the spring can be found by equating the kinetic energy of the system to the potential energy of the spring at maximum elongation.
Step 1: Find the initial kinetic energy of the system. Since both blocks are moving with the same speed in opposite directions, the total kinetic energy (K.E.) is given by:
K.E. = 1/2 * m * v^2 + 1/2 * m * v^2 = m * v^2
Step 2: At maximum elongation, the blocks come to rest momentarily, so all the kinetic energy is converted into potential energy stored in the spring. The potential energy (P.E.) of a spring is given by:
P.E. = 1/2 * k * x^2
where x is the elongation of the spring.
Step 3: Equate the kinetic energy to the potential energy to find the maximum elongation:
m * v^2 = 1/2 * k * x^2
Solving for x gives:
x = sqrt((2 * m * v^2) / k)
So, the maximum elongation of the spring is sqrt((2 * m * v^2) / k).
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