A point particle of mass 0.5 ,kg is moving al | vedclass

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(B) The particle is at the origin $(x=0)$ and is projected towards the right. To escape to infinity on the right side,it must overcome the potential b

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$2 \,m/s$

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(B) The particle is at the origin $(x=0)$ and is projected towards the right. To escape to infinity on the right side,it must overcome the potential barrier of $4 \,J$.Using the law of conservation of energy: $K_i + V_i = K_f + V_f$.At the origin,$V_i = 0$ and $K_i = \frac{1}{2}mv^2$.To just cross the barrier,the final kinetic energy at the peak $(V_f = 4 \,J)$ must be at least $0$.$\frac{1}{2} 	imes 0.5 	imes v^2 = 4 \Rightarrow 0.25 	imes v^2 = 4 \Rightarrow v^2 = 16 \Rightarrow v = 4 \,m/s$.However,if the particle does not have enough energy to cross the right barrier,it will be reflected back towards the left. On the left side,there is a smaller potential barrier of $1 \,J$.If the particle is reflected,it moves towards the left and must overcome the barrier of $1 \,J$ to escape to infinity on the left side.For this,the initial kinetic energy must be at least $1 \,J$:$\frac{1}{2} 	imes 0.5 	imes v^2 = 1 \Rightarrow 0.25 	imes v^2 = 1 \Rightarrow v^2 = 4 \Rightarrow v = 2 \,m/s$.Since $2 \,m/s < 4 \,m/s$,the minimum speed required for the particle to escape to infinity (by going over the left barrier after reflection) is $2 \,m/s$.

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