A mass of 100,g strikes the wall with a speed | vedclass

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(C) The force applied on the mass is given by the rate of change of momentum: $\vec{F} = \frac{\Delta \vec{P}}{\Delta t}$.Let the wall be along the $y

Vedclass · Accepted Answer

$250\sqrt{3}\,N$ to the left

Vedclass · Answer

(C) The force applied on the mass is given by the rate of change of momentum: $\vec{F} = \frac{\Delta \vec{P}}{\Delta t}$.Let the wall be along the $y$-axis. The initial velocity vector is $\vec{v}_i = v \sin 	heta \hat{i} + v \cos 	heta \hat{j}$.The final velocity vector is $\vec{v}_f = -v \sin 	heta \hat{i} + v \cos 	heta \hat{j}$.The change in momentum is $\Delta \vec{P} = m(\vec{v}_f - \vec{v}_i) = m(-2v \sin 	heta \hat{i}) = -2mv \sin 	heta \hat{i}$.Given $m = 0.1\,kg$,$v = 5\,m/s$,$	heta = 60^\circ$,and $\Delta t = 2 	imes 10^{-3}\,s$.Substituting the values: $\Delta P_x = -2 	imes 0.1 	imes 5 	imes \sin(60^\circ) = -1 	imes \frac{\sqrt{3}}{2} = -0.5\sqrt{3}\,kg\cdot m/s$.The force $\vec{F} = \frac{-0.5\sqrt{3}}{2 	imes 10^{-3}} = -250\sqrt{3}\,N$.The negative sign indicates that the force exerted by the wall on the mass is directed to the left.

$A$ mass of $100\,g$ strikes the wall with a speed of $5\,m/s$ at an angle as shown in the figure and it rebounds with the same speed. If the contact time is $2 \times 10^{-3}\,s$,what is the force applied on the mass by the wall?

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