Question

A mass of $0.5kg$ moving with a speed of $1.5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$ on a horizontal smooth surface collides with a nearly weightless spring of force constant, $k=50N/m$. The maximum compression of the spring would be

Answer 1

Step 1. Given data:

Mass = $0.5kg$

Speed = $1.5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Force constant of spring , $k=50N/m$

Step 2, Formula used:

From law of conservation of energy, we know that

$$\begin{gathered} \left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)m{v}^2=\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)k.x^2 \\ m{v}^2=k.x^2 \end{gathered}$$

Step 3. Finding maximum compression

On substituting the given values in the above equation we get

$$\begin{gathered} \left(0.5\right){\left(1.5\right)}^2=\left(50\right).x^2 \\ x=0.15m \end{gathered}$$

Hence, the maximum compression of the spring would be $0.15m.$