Question

A $3\text{kg}$ block collides with a massless spring of spring constant $110\text{N/m}$ attached to a wall. The speed of the block was observed to be $1.4\text{m/s}$ at the moment of collision. The acceleration due to gravity is $9.8{\text{m/s}}^2$. How far does the spring compress if the horizontal surface on which the mass moves is frictionless?

• A. $0.23\text{m}$
• B. $0.41\text{m}$
• C. $2.5\text{m}$
• D. $3.2\text{m}$

Answer 1

The correct option is A $0.23\text{m}$

Given, mass of block, $m=3\text{kg}$
and spring constant $k=110\text{N/m}$
Initial velocity of block $v_i=1.4\text{m/s}$
No initial elongation or compression in spring
$\therefore x_i=0\text{m}$
Final velocity of block $v_f=0\text{m/s}$
(at maximum compression, block comes to rest)

Applying energy conservation from the point where the block just collides with the spring, till the block stops, when spring compresses to maximum:
$K{E}_i+U_i=K{E}_f+U_f$
$\frac{1}{2}m{v}_i^2+\frac{1}{2}k{x}_i^2=\frac{1}{2}m{v}_f^2+\frac{1}{2}k{x}_f^2$
$\frac{1}{2}m{v}_i^2+\frac{1}{2}k(0{)}^2=\frac{1}{2}m(0{)}^2+\frac{1}{2}k{x}_f^2$

Or, $m{v}_i^2=k{x}_f^2$
$\Rightarrow x_f=\sqrt{\frac{m{v}_i^2}{k}}$
Putting the values of $m,v_i$ and $k$ gives
$x_f=\sqrt{\frac{3\times {1.4}^2}{110}}=\frac{1}{10}\times \sqrt{\frac{5.88}{1.1}}$
$\therefore x_f=0.23\text{m}$