As shown in the figure, a block of mass 2.0 kg travels at a...

Let $m_1 = 2.0 \text{ kg}$, $u_1 = 1.0 \text{ m/s}$, $m_2 = 2.0 \text{ kg}$, $u_2 = 0 \text{ m/s}$, and $k = 100 \text{ N/m}$.

At the instant of maximum compression, both blocks move with the same common velocity $v$.

By conservation of linear momentum:

$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$

$2.0 \times 1.0 + 0 = (2.0 + 2.0) v$

$2.0 = 4.0 v \Rightarrow v = 0.5 \text{ m/s}$

By conservation of mechanical energy, the loss in kinetic energy is stored as the potential energy of the spring:

$\dfrac{1}{2} m_1 u_1^2 = \dfrac{1}{2} (m_1 + m_2) v^2 + \dfrac{1}{2} k x^2$

where $x$ is the maximum compression of the spring.

$\dfrac{1}{2} \times 2.0 \times (1.0)^2 = \dfrac{1}{2} \times 4.0 \times (0.5)^2 + \dfrac{1}{2} \times 100 \times x^2$

$1.0 = 2.0 \times 0.25 + 50 x^2$

$1.0 = 0.5 + 50 x^2$

$50 x^2 = 0.5$

$x^2 = \dfrac{0.5}{50} = \dfrac{1}{100} \text{ m}^2$

$x = 0.1 \text{ m} = 10 \text{ cm}$