Question

Two masses $m_1=2kg$ and $m_2=5kg$ are moving on a frictionless surface with velocities 10 m/s and 3 m/s respectively. $m_2$ is ahead of $m_1$. An ideal spring of spring constant $k=1120N/m$ is attached on the back side of $m_2$.The maximum compression of the spring will be

• A. $0.51m$
• B. $0.062m$
• C. $0.37m$
• D. $0.72m$

Answer 1

Answer: (C)

$0.37m$

At the time of maximum compression of spring velocity of both blocks will same.

By momentum conservation,

$m_1{v}_1+m_2{v}_2=(m_1+m_2)u$

$2\times 10+5\times 3=(2+5)u$

$25=7u$

$u=\frac{25}{7}m/s=3.57m/s$

By energy conservation,

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

$\frac{1}{2}\times 2\times 100+\frac{1}{2}\times 5\times 9=\frac{1}{2}\times 2\times (3.57{)}^2+\frac{1}{2}\times 5\times (3.57{)}^2+\frac{1}{2}k{x}^2$

$122.5=44.6+\frac{1}{2}k{x}^2$

$\frac{1}{2}k{x}^2=77.9$

$x^2=\frac{2\left(77.9\right)}{k}=\frac{77.9}{1120}\times 2$

$x=0.37$