Question

A body of mass \(1~kg\) falls freely from a height of \(100~m\) on a platform of mass \(3~kg\) which is mounted on a spring having spring constant \(K=1.25\times 10^6 N/m\).The body sticks to the platform and the spring's maximum comperssion is found to be \(x\). Given that \(g=10 m/s^2\), the value of \(x\) will be close to

Answer 1

Step 1: Find the velocity of \(1~kg\) block just before it collides.

As we know,
\(\text{Intial velocity}\) of \(1~kg\) \(\text{mass}\) \(u=0\)
Hence, by using \(3^{rd}\) equation of motion,
\(v^2-u^2=2gh\)
Therefore, by putting the value we get
velocity of \(1~kg\) block just before it collides with \(3~kg \) block
\(=\sqrt{2gh}=\sqrt{2000}~m/s\)

Step 2: Find the value of \(v\).
Applying momentum conversation just before and just after collision
\(1\times\sqrt{2000}=4v\)
\(v=\dfrac{\sqrt{2000}}{4}m/s\)

Step 3: Find the value of maximum comperssion\((x)\)
Intial comperssion of spring
\(kx_0=F\)
As we know , \(F=mg=3\times10=30~N\)
Therefore, by putting values in above equation,
\(1.25\times10^6x_0=30\Rightarrow x_0\approx 0\)
As we know,work energy theorem,
\(\text{Work done by gravity }+\text{work done by the spring} = \text{Change in kinetic energy} \)
Hence,
\(W_g+W_{sp}=\Delta K.E\)
by putting the values, we get
\(40\times x+\dfrac{1}{2}\times 1.25\times 10^6(0^2-x^2)\)
\(x=0-\dfrac{1}{2}\times4\times v^2\)
By solving we get,
\(x=4cm\)