A thin uniform bar lies on a frictionless hor | vedclass

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(A) Let the mass of the bar be $M = 0.300 \, kg$ and the mass of each particle be $m = 0.100 \, kg$.Since the surface is frictionless and there are no

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$3$

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(A) Let the mass of the bar be $M = 0.300 \, kg$ and the mass of each particle be $m = 0.100 \, kg$.Since the surface is frictionless and there are no external horizontal forces acting on the system (bar + two particles),the linear momentum of the system is conserved in the direction of the particles' motion.Let $u_1 = 10 \, m/s$ and $u_2 = 5 \, m/s$ be the velocities of the two particles before the collision.Let $V_{cm}$ be the velocity of the centre of mass of the bar after the collision. Note that since the particles strike the ends simultaneously,the system's total momentum is conserved.According to the law of conservation of linear momentum:$m u_1 + m u_2 = (M + 2m) V_{cm}$Substituting the given values:$0.100(10) + 0.100(5) = (0.300 + 2 	imes 0.100) V_{cm}$$1.0 + 0.5 = (0.300 + 0.200) V_{cm}$$1.5 = 0.5 V_{cm}$$V_{cm} = \frac{1.5}{0.5} = 3 \, m/s$Thus,the velocity of the centre of mass of the bar after the collision is $3 \, m/s$.

$(a)$ $\frac{mg}{4} < F < \frac{mg}{2}$	$(i)$ Cube will move up
$(b)$ $F > \frac{mg}{2}$	$(ii)$ Cube will not exhibit motion
$(c)$ $F > mg$	$(iii)$ Cube will begin to rotate about $A$
$(d)$ $F = \frac{mg}{4}$	$(iv)$ Normal reaction effectively at $a/3$ from $A$,no motion

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