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Question
A block of mass 2.0 kg is moving on a frictionless horizontal surface with a velocity of 1.0 m/s (In the following figure) towards another block of equal mass kept at rest. The spring constant of the spring fixed at one end is 100 N/m. Find the maximum compression of the spring.
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Solution
Given,
Mass of each block, MA = MB = 2 kg
Initial velocities of block A, Va = 1 m/s
Initial velocity of block B, Vb = 0
Spring constant of the spring = 100 N/m
Block A strikes the spring with a velocity of 1 m/s.
After the collision, it's velocity decreases continuously. At an instant the whole system (Block A + the compound spring + Block B) moves together with a common velocity V (say).
Using the law of conservation of energy, we get:
\[\left( \frac{1}{2} \right) M_A V_A^2 + \left( \frac{1}{2} \right) M_B V_B^2 = \left( \frac{1}{2} \right) M_A V^2 + \left( \frac{1}{2} \right) M_B V^2 + \left( \frac{1}{2} \right)k x^2 \]
\[\]
\[\left( \frac{1}{2} \right) \times 2(1 )^2 + 0 = \left( \frac{1}{2} \right) + \left( \frac{1}{2} \right) \times v^2 + \left( \frac{1}{2} \right) x^2 \times 100\]
(where x is the maximum compression of the spring)
⇒ 1 − 2v2 = 50x2 ...(1)
As there is no external force acting in the horizontal direction, the momentum is conserved.
\[\Rightarrow M_A V_A + M_B V_B = ( M_A + M_B )V\]
\[ \Rightarrow 2 \times 1 = 4 \times V\]
\[ \Rightarrow V = \left( \frac{1}{2} \right)\text{m/s} . . . (2)\]
\[\text{ Susbstituting this value of V in equation (1), we get: }\]
\[ 1 = 2 \times \left( \frac{1}{4} \right) + 50 x^2 \]
\[ \Rightarrow \frac{1}{4} = 50 x^2 \]
\[ \Rightarrow x^2 = \frac{1}{100} \]
\[ \Rightarrow x = \frac{1}{10}m\]
\[ \Rightarrow x = 10 \text{cm}\]
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