Question

A block of mass $m_1=2kg$ slides along a frictionless table with a speed of 10m/s. Directly in front of it, and moving in same direction is a block of mass $m_2=5kg$ moving at a speed of 3 m/s. A massless spring with a constant of k = 1120 N/m is attached to the back side of $m_2$ as shown. When the block collides, what is the maximum compression in the spring?

• A. 0.25 m
• B. 0.50 m
• C. 0.125 m
• D. 0.75 m

Answer 1

The correct option is A

0.25 m

When the block $m_1$ strikes $m_2$, the spring beings to get compressed and $m_2$ gains speed. At the instant when $m_2$ and $m_1$ have equal velocities, the compression in the spring is maximum.

Let v = common velocity of blocks, now applying momentum conservation:

$m_1{u}_1+m_2{u}_2=m_{1}v+m_{2}v$ $\Rightarrow$ v = $\frac{m_1{u}_1+m_2{u}_2}{m_1+m_2}=5m/s.$

Using conversation of energy: Loss in K.E = gain in elastic PE

$\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2-(\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2)=\frac{1}{2}k{x}^2$

$200+45-7\left(25\right)=1120x^2\Rightarrow x=0.25m$.