A-particle-of-mass-m-moving-with-speed-u-collides-perfectly-inelastically-with-a-sphere-of-radius-R-and-same-mass-at-rest-at-an-impact-parameter-d-Find-a-Angle-between-their-final-velocities-b-

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Question Number 25091 by Tinkutara last updated on 03/Dec/17

$$Aparticleofmassmmovingwith$$$$speeducollidesperfectlyinelastically$$$$withasphereofradiusRandsame$$$$mass,atrest,atanimpactparameter$$$$d.Find$$$$\left(a\right)Anglebetweentheirfinalvelocities$$$$\left(b\right)Magnitudeoftheirfinal$$$$velocities$$

Answered by ajfour last updated on 03/Dec/17

Commented by ajfour last updated on 04/Dec/17

$$conservinglinearmomentum$$$$mu=2mv+m\omega R\sin \theta$$$$and\sin \theta =d/R,so$$$$2v+\omega d=u\dots \left(i\right)$$$$conservingangularmomentum$$$$aboutgroundpointG:$$$$mu(R+d)=mvR+\frac{2}{5}{mR}^2\omega +$$$$m(v+\omega R\sin \theta )(R+d)+m\omega {\left(R\cos \theta \right)}^2$$$$\dots .\left(ii\right)$$$$andas{\left(R\cos \theta \right)}^2=(R+d)(R-d),$$$$alsoR\sin \theta =d$$$$\Rightarrow u(R+d)=vR+\frac{2}{5}\omega R^2+$$$$(v+\omega d)(R+d)+\omega (R^2-d^2)$$$$from\left(i\right):v+\omega d=u-v,so$$$$vd=\omega (\frac{7}{5}{R}^2-d^2)$$$$andfrom\left(i\right):2\text{boldsymbol}v+\text{boldsymbol}\omega d=\text{boldsymbol}u$$$$\Rightarrow 2\omega (\frac{7}{5}{R}^2-d^2)+\omega d^2=ud$$$$\omega =\frac{ud}{(\frac{14}{5}{R}^2+d^2)};\text{boldsymbol}v=\frac{\text{boldsymbol}u(\frac{7}{5}\text{boldsymbol}{R}^2-d^2)}{(\frac{14}{5}{R}^2+d^2)}$$$$\tan \varphi =\frac{\omega R\cos \theta }{v+\omega d}=\frac{d\sqrt{R^2-d^2}}{\left(vd/\omega \right)+d^2}$$$$\tan \text{boldsymbol}\varphi =\frac{5\text{boldsymbol}d\sqrt{\text{boldsymbol}{R}^2-\text{boldsymbol}{d}^2}}{7\text{boldsymbol}{R}^2}.$$$$V=\frac{\omega R\cos \theta }{\sin \varphi }=\omega \left[\frac{\sqrt{R^2-d^2}}{\sin \varphi }\right].$$

Commented by ajfour last updated on 03/Dec/17

$$\mathrm{\angle }betweentheirfinalvelocities$$$$is\text{boldsymbol}\varphi .$$$$finalvelocityofsphere\text{boldsymbol}v.$$$$Finalvelocityofparticleis\text{boldsymbol}V.$$

Commented by Tinkutara last updated on 05/Dec/17

$$Answeris:$$$$v_1=\frac{ud}{R},v_2=-v_3=\frac{u\sqrt{R^2-d^2}}{2R}$$

Commented by ajfour last updated on 05/Dec/17

$$whichansweriscorrectfor$$$$d=0,andevend=R?Please$$$$checkorithinkihave$$$$misunderstoodthequestion.$$$$mrWsircanhelpifhefinds$$$$timeandnoticesmyrequest.$$

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