Question

A block of mass 200 g is suspended through a vertical spring. The spring is stretched by 1.0 cm when the block is in equilibrium. A particle of mass 120 g is dropped on the block from a height of 45 cm. The particle sticks to the block after the impact. Find the maximum extension of the spring.
Figure

Answer 1

It is given that:
Mass of block, M = 200 g = 0.20 kg
Mass of the particle, m = 120 gm = 0.12 kg
Height of the particle, h = 45 cm = 0.45 m



According to question, as the block attains equilibrium, the spring is stretched by a distance, x = 1.00 cm = 0.01 m.

i.e. M × g = K × x

⇒ 0.2 × g = K × x
⇒ 2 = K × 0.01
⇒ K = 200 N/m

The velocity with which the particle m strikes M is given by,
$$\begin{gathered} u=\sqrt{2gh} \\ u=\sqrt{2\times 10\times 0.45} \\ =\sqrt{9}=3\mathrm{m}/\mathrm{s} \end{gathered}$$

After the collision, let the velocity of the particle and the block be V.
According to law of conservation of momentum, we can write:
mu = (m + M)V

Solving for V , we get:
$V=\frac{0.12\times 3}{0.32}=\frac{9}{8}\mathrm{m}/\mathrm{s}$

Let the spring be stretched through an extra deflection of δ.

On applying the law of conservation of energy, we can write:
Initial energy of the system before collision = Final energy of the system

$\Rightarrow \frac{1}{2}m{u}^2+\frac{1}{2}K{x}^2=\frac{1}{2}\left(m+M\right)V^2+\frac{1}{2}K{\left(x+\delta \right)}^2$

Substituting appropriate values in the above equation, we get:
$\left(\frac{1}{2}\right)\times 0.12\times 9+\left(\frac{1}{2}\right)\times 200\times (0.01{)}^2=\left(\frac{1}{2}\right)0.32\times \left(\frac{81}{64}\right)+\left(\frac{1}{2}\right)\times 200\times (\delta +0.1{)}^2$

On solving the above equation, we get:
δ = 0.061
m = 6.1 cm