Physics - W.E.P. Question with Solution | TestHub

With $m=60\text{g}=0.06\text{kg}$ and $g=10$ m/s², $mg=0.6$ N. The lower ball just leaves the ground when the tension equals $mg$; the thread is then 1 m long, i.e. the extension is $x_0=1-0.4=0.6$ m, so

$$k{x}_0=mg\Rightarrow k=\frac{0.6}{0.6}=1\text{N/m}$$

While the upper ball is raised slowly, the external force balances gravity and the thread tension: $F_{ext}=kx+mg$ ($x$: extension at any instant). Hence

$$W_{ext}={\int }_0^{x_0}\left(kx+mg\right)dx=\frac{1}{2}k{x}_0^2+mg{x}_0=\frac{1}{2}(1){\left(0.6\right)}^2+(0.6)(0.6)=0.18+0.36=0.54\text{J}\approx 0.53\text{J}$$

so (A) is correct and (B) is false.

On release, the stretched thread (extension 0.6 m) first pulls the upper ball down; while the thread contracts from extension $x_0$ to zero, both gravity and the thread tension do positive work on it:

$$\frac{1}{2}m{v}_1^2=mg{x}_0+\frac{1}{2}k{x}_0^2=0.36+0.18=0.54\text{J}\Rightarrow v_1=\sqrt{\frac{2\times 0.54}{0.06}}=\sqrt{18}=4.24\text{m/s}$$

The ball then falls freely through the natural length $h=0.4$ m before hitting the lower ball:

$$v^2=v_1^2+2gh=18+2\times 10\times 0.4=26\Rightarrow v=\sqrt{26}\approx 5.1\text{m/s}$$

so (C) is correct. For the time: during contraction the motion is simple harmonic about the point where $kx=mg$ with $\omega =\sqrt{k/m}=\sqrt{1/0.06}=4.08$ rad/s; measuring the fall $y$ from the release point, $y=1.2(1-\text{cos}\omega t)$, so $y=0.6$ m when $\text{cos}\omega t_1=1/2$, i.e. $t_1=\frac{\pi /3}{4.08}=0.257$ s. The remaining free fall of 0.4 m takes $t_2$:

$$0.4=4.24t_2+\frac{1}{2}(10)t_2^2\Rightarrow t_2=0.086\text{s}$$

$$\therefore t=t_1+t_2=0.257+0.086\approx 0.34\text{s}$$

so (D) is also correct.