First slide

Mechanical energy conservation

Question

A mass of 0.5 kg moving with a speed of 1.5 m/s on a horizontal smooth surface collides with a nearly weightless spring of force constant k = 50 N/m. The maximum compression of the spring would be:

Easy

A

0.15 m
B

0.12 m
C

1.5 m
D

0.5 m

Solution

By the law of conservation of energy,
$\begin{array}{ll} i.e., ΔK + ΔU = 0 \\ (0 - \frac{1}{2} {mv}^{2}) + (\frac{1}{2} {kx}^{2} - 0) = 0 \end{array}$
$\begin{aligned} \frac{1}{2} {mv}^{2} = \frac{1}{2} {kx}^{2} \\ \Rightarrow x^{2} = \frac{{mv}^{2}}{k} \end{aligned}$
$\begin{aligned} \Rightarrow x = \sqrt{(\frac{{mv}^{2}}{k})} \\ \Rightarrow x = \sqrt{(\frac{0 .5 \times 1 .5 \times 1 .5}{50})} \end{aligned}$
$∴= 0.15 m$

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