Question Number 29254 by Tinkutara last updated on 06/Feb/18
Commented by Tinkutara last updated on 06/Feb/18
Find the exact velocity vectors with x and y components. e=1/3 Theta=60° m=3 kg M=8 kg L=6 m u=25 m/s Tell me your final answers so that I can check whether its correct or not.
Answered by mrW2 last updated on 06/Feb/18
$${u}_{\mathrm{1}{x}} ,{u}_{\mathrm{1}{y}} ={velocity}\:{of}\:{particle} \\ $$$${u}_{\mathrm{2}{x}} ,{u}_{\mathrm{2}{y}} ={velocity}\:{of}\:{rod} \\ $$$${u}_{\mathrm{1}{y}} −{u}_{\mathrm{2}{y}} ={u}\:\mathrm{sin}\:\theta \\ $$$${u}_{\mathrm{2}{x}} −{u}_{\mathrm{1}{x}} ={eu}\:\mathrm{cos}\:\theta \\ $$$$ \\ $$$${mu}_{\mathrm{1}{x}} +{Mu}_{\mathrm{2}{x}} ={mu}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow{mu}_{\mathrm{1}{x}} +{M}\left({eu}\:\mathrm{cos}\:\theta+{u}_{\mathrm{1}{x}} \right)={mu}\:\mathrm{cos}\:\theta \\ $$$$\Rightarrow\left({M}+{m}\right){u}_{\mathrm{1}{x}} ={u}\:\mathrm{cos}\:\theta\left({m}−{Me}\right) \\ $$$$\Rightarrow{u}_{\mathrm{1}{x}} =\frac{\left({m}−{Me}\right)\:\mathrm{cos}\:\theta\:{u}}{{M}+{m}} \\ $$$$\Rightarrow{u}_{\mathrm{2}{x}} =\frac{\left(\mathrm{1}+{e}\right){m}\:\mathrm{cos}\:\theta}{{M}+{m}}\:{u} \\ $$$$ \\ $$$${mu}_{\mathrm{1}{y}} +{Mu}_{\mathrm{2}{y}} ={mu}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow{mu}_{\mathrm{1}{y}} +{M}\left({u}_{\mathrm{1}{y}} −{u}\:\mathrm{sin}\:\theta\right)={mu}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow{u}_{\mathrm{1}{y}} ={u}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow{u}_{\mathrm{2}{y}} =\mathrm{0} \\ $$$$ \\ $$$${e}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\theta=\mathrm{60}° \\ $$$${m}=\mathrm{3}\:{kg} \\ $$$${M}=\mathrm{8}\:{kg} \\ $$$${u}=\mathrm{25}\:{m}/{s} \\ $$$$\Rightarrow{u}_{\mathrm{1}{x}} =\frac{\left(\mathrm{3}−\frac{\mathrm{8}}{\mathrm{3}}\right)×\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{25}}{\mathrm{8}+\mathrm{3}}=\frac{\mathrm{25}}{\mathrm{66}}\approx\mathrm{0}.\mathrm{38}\:{m}/{s} \\ $$$$\Rightarrow{u}_{\mathrm{1}{y}} =\mathrm{25}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\approx\mathrm{21}.\mathrm{7}\:{m}/{s} \\ $$$$ \\ $$$$\Rightarrow{u}_{\mathrm{2}{x}} =\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\right)×\mathrm{3}×\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{8}+\mathrm{3}}×\mathrm{25}=\frac{\mathrm{50}}{\mathrm{11}}\approx\mathrm{4}.\mathrm{5}\:{m}/{s} \\ $$$$\Rightarrow{u}_{\mathrm{2}{y}} =\mathrm{0} \\ $$
Commented by Tinkutara last updated on 06/Feb/18
Why momentum is conserved in x direction?
Commented by mrW2 last updated on 06/Feb/18
I don't understand your doubt. The momentum should be conserved in every direction, so why not in x direction?
Commented by Tinkutara last updated on 06/Feb/18
But momentum is not conserved in that direction in which impulsive forces act so that's why I am asking this.
Commented by mrW2 last updated on 06/Feb/18
$${In}\:{our}\:{case}\:{no}\:{force}\:{acts}\:{both}\:{in}\:{x}\:{and} \\ $$$${in}\:{y}\:{direction}. \\ $$
Commented by Tinkutara last updated on 07/Feb/18
Thank you very much Sir! I got the answer.