Question Number 27677 by Tinkutara last updated on 12/Jan/18
Commented by Tinkutara last updated on 12/Jan/18
Commented by ajfour last updated on 14/Jan/18
Commented by ajfour last updated on 14/Jan/18
$${before}\:{collision}\:{at}\:{B} \\ $$$${v}=\sqrt{\mathrm{2}×\mathrm{10}×\mathrm{3}}\:=\sqrt{\mathrm{60}}\:{m}/{s} \\ $$$${after}\:{inelastic}\:{collision}\:{at}\:{B} \\ $$$${v}=\sqrt{\mathrm{60}}\:\mathrm{cos}\:\mathrm{30}°\:=\sqrt{\mathrm{60}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\sqrt{\mathrm{45}}{m}/{s}. \\ $$$${just}\:{before}\:{leaving}\:{second}\:{incline} \\ $$$${speed}\:{at}\:{C}\:: \\ $$$${v}_{{C}} ^{\mathrm{2}} \:=\:\sqrt{\mathrm{45}+\mathrm{2}×\mathrm{10}×\mathrm{3}}\:=\sqrt{\mathrm{105}}\:{m}/{s} \\ $$$${if}\:{collision}\:{were}\:{elastic} \\ $$$${after}\:{collision}\:{at}\:{B}, \\ $$$${v}_{{y}} =\:\sqrt{\mathrm{15}}\mathrm{cos}\:\mathrm{30}°−\sqrt{\mathrm{45}}\mathrm{sin}\:\mathrm{30}° \\ $$$$\:\:\:\:\:=\frac{\sqrt{\mathrm{15}}×\sqrt{\mathrm{3}}}{\mathrm{2}}−\sqrt{\mathrm{45}}×\frac{\mathrm{1}}{\mathrm{2}}\:=\mathrm{0}\:. \\ $$
Commented by Tinkutara last updated on 14/Jan/18
Thanks very much!
Answered by ajfour last updated on 14/Jan/18
$$\mathrm{1}\rightarrow\left(\mathrm{2}\right) \\ $$$$\mathrm{2}\rightarrow\left(\mathrm{2}\right) \\ $$$$\mathrm{3}\rightarrow\left(\mathrm{3}\right) \\ $$
Commented by Tinkutara last updated on 14/Jan/18
Thank you! I solved it.