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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

$$u_{1x},u_{1y}=velocityofparticle$$$$u_{2x},u_{2y}=velocityofrod$$$$u_{1y}-u_{2y}=u\sin \theta$$$$u_{2x}-u_{1x}=eu\cos \theta$$$$$$$${mu}_{1x}+{Mu}_{2x}=mu\cos \theta$$$$\Rightarrow {mu}_{1x}+M(eu\cos \theta +u_{1x})=mu\cos \theta$$$$\Rightarrow (M+m)u_{1x}=u\cos \theta (m-Me)$$$$\Rightarrow u_{1x}=\frac{(m-Me)\cos \theta u}{M+m}$$$$\Rightarrow u_{2x}=\frac{(1+e)m\cos \theta }{M+m}u$$$$$$$${mu}_{1y}+{Mu}_{2y}=mu\sin \theta$$$$\Rightarrow {mu}_{1y}+M(u_{1y}-u\sin \theta )=mu\sin \theta$$$$\Rightarrow u_{1y}=u\sin \theta$$$$\Rightarrow u_{2y}=0$$$$$$$$e=\frac{1}{3}$$$$\theta =60°$$$$m=3kg$$$$M=8kg$$$$u=25m/s$$$$\Rightarrow u_{1x}=\frac{(3-\frac{8}{3})\times \frac{1}{2}\times 25}{8+3}=\frac{25}{66}\approx 0.38m/s$$$$\Rightarrow u_{1y}=25\times \frac{\sqrt{3}}{2}\approx 21.7m/s$$$$$$$$\Rightarrow u_{2x}=\frac{(1+\frac{1}{3})\times 3\times \frac{1}{2}}{8+3}\times 25=\frac{50}{11}\approx 4.5m/s$$$$\Rightarrow u_{2y}=0$$

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

$$Inourcasenoforceactsbothinxand$$$$inydirection.$$

Commented by Tinkutara last updated on 07/Feb/18

Thank you very much Sir! I got the answer.

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