A body of mass M moving at speed V_{0} collid | vedclass

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(C) Given $	heta_{1} = 	heta_{2} = 	heta$.From the conservation of linear momentum:In the $x$-direction: $M V_{0} = M V_{1} \cos 	heta + m V_{2} \

Vedclass · Accepted Answer

$3$

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(C) Given $	heta_{1} = 	heta_{2} = 	heta$.From the conservation of linear momentum:In the $x$-direction: $M V_{0} = M V_{1} \cos 	heta + m V_{2} \cos 	heta$In the $y$-direction: $0 = M V_{1} \sin 	heta - m V_{2} \sin 	heta$From the $y$-direction equation,we get $M V_{1} = m V_{2}$,so $V_{2} = \frac{M V_{1}}{m}$.Substituting this into the $x$-direction equation: $M V_{0} = (M V_{1} + m \cdot \frac{M V_{1}}{m}) \cos 	heta = 2 M V_{1} \cos 	heta$,which gives $V_{0} = 2 V_{1} \cos 	heta$.From the conservation of kinetic energy in an elastic collision:$\frac{1}{2} M V_{0}^{2} = \frac{1}{2} M V_{1}^{2} + \frac{1}{2} m V_{2}^{2}$Substituting $V_{0} = 2 V_{1} \cos 	heta$ and $V_{2} = \frac{M V_{1}}{m}$:$M (4 V_{1}^{2} \cos^{2} 	heta) = M V_{1}^{2} + m (\frac{M V_{1}}{m})^{2}$$4 M \cos^{2} 	heta = M + \frac{M^{2}}{m}$Dividing by $M$: $4 \cos^{2} 	heta = 1 + \frac{M}{m}$Since $\cos^{2} 	heta \leq 1$,we have $1 + \frac{M}{m} \leq 4$,which implies $\frac{M}{m} \leq 3$.Thus,the largest possible value of the ratio $M/m$ is $3$.

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