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.

Answer 1

Given,
Mass of each block, M_A = M_B = 2 kg 
Initial velocities of block A, V_a = 1 m/s
Initial velocity of block B, V_b = 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.

\[\Rightarrow M_A V_A + M_B V_B = ( M_A + M_B )V\]

\[ \Rightarrow 2 \times 1 = 4 \times V\]

\[ \Rightarrow V = \left( \frac{1}{2} \right)\text{m/s} . . . (2)\]

\[\text{ Susbstituting this value of V in equation (1), we get: }\]

\[ 1 = 2 \times \left( \frac{1}{4} \right) + 50 x^2 \]

\[ \Rightarrow \frac{1}{4} = 50 x^2 \]

\[ \Rightarrow x^2 = \frac{1}{100} \]

\[ \Rightarrow x = \frac{1}{10}m\]

\[ \Rightarrow x = 10 \text{cm}\]