Question

A block of mass 2.0 kg is moving on a frictionless horizontal surface with a velocity of 1.0 m/s 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
Figure

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:

$$\begin{gathered} \left(\frac{1}{2}\right)M_A{V}_A^2+\left(\frac{1}{2}\right)M_B{V}_B^2=\left(\frac{1}{2}\right)M_A{V}^2+\left(\frac{1}{2}\right)M_B{V}^2+\left(\frac{1}{2}\right)k{x}^2 \\ \left(\frac{1}{2}\right)\times 2(1{)}^2+0=\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)\times v^2+\left(\frac{1}{2}\right)x^2\times 100 \end{gathered}$$
(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.

$$\begin{gathered} \Rightarrow M_A{V}_A+M_B{V}_B=(M_A+M_B)V \\ \Rightarrow 2\times 1=4\times V \\ \Rightarrow V=\left(\frac{1}{2}\right)\mathrm{m}/\mathrm{s}...\left(2\right) \\ \mathrm{Susbstituting}\mathrm{this}\mathrm{value}\mathrm{of}\mathrm{V}\mathrm{in}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}: \\ 1=2\times \left(\frac{1}{4}\right)+50x^2 \\ \Rightarrow \frac{1}{4}=50x^2\mathit{} \\ \Rightarrow x^2=\frac{1}{100} \\ \Rightarrow x=\frac{1}{10}\mathrm{m} \\ \Rightarrow x=10\mathrm{cm} \end{gathered}$$