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.Take g=10 $m/s^2$.

Answer 1

Mass of block M=200 g=0.20 kg

Mass of the particle,m=120gm=0.12 kg

In the equilibrium condition,the spring is stretched by a distance,

x = 1.00 cm = 0.01 m

By balancing the forces in the equilibrium position we have,

$Mg=kx$

$\Rightarrow 0.2\times g=k\times x$

$\Rightarrow 2=k\times 0.01$

$\Rightarrow k=200N/m$

The velocity with which the particle m will strike M is given by,

$u=\sqrt{2gh}=\sqrt{2\times 10\times 0.45}$

$=\sqrt{9}=3m/sec$

So,after the collision, the velocity of the particle and the block is

$v=\frac{0.12\times 3}{0.32}=\frac{9}{8}m/sec$

Let the spring be stretched through an extra deflection of $\delta$.

$0-\left(\frac{1}{2}\right)\times 0.32\times \left(\frac{81}{64}\right)$

$=0.32\times 10\times \delta -\left(\frac{1}{2}\right)\times 200\times (\delta +0.1)62-\left(\frac{1}{2}\right)\times 200\times (0.01{)}^2$

Solving the above equation,we can get $\delta$=0.045m

so $x_{max}=1cm+\delta$

$x_{max}=1+4.5$ = 5.5 cm