Physics - Center of Mass Question with Solution | TestHub

$V_{cm}=\frac{0.25\times 1}{1.25}=0.2\mathrm{m}/\mathrm{s}$ and friction eventually brings the block to rest on the trolley - (A) true. Relative deceleration$=1+0.25=1.25\mathrm{m}/{\mathrm{s}}^2$; relative rest needs$\frac{1}{2.5}=0.4\mathrm{m}>0.3\mathrm{m}$, so the block hits the far wall once with relative speed$\sqrt{1-0.75}=0.5\mathrm{m}/\mathrm{s}$. After the elastic bounce it covers only$\frac{0.25}{2.5}=0.1\mathrm{m}$ : stops 0.2 m from the rear wall, after just one wall collision - (B), (C) false. After the bounce the friction on the trolley reverses, so the trolley slows down - (D) false. Answer: (A)