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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$2 \,m/s$

Vedclass · Answer

(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 $2 \,m/s$.

Similar Questions

The potential energy of a diatomic molecule is given by $U = \frac{A}{r^{12}} - \frac{B}{r^6}$,where $A$ and $B$ are positive constants. The distance $r$ between them at equilibrium is:

The potential energy of a particle in a force field is $U(r) = \frac{A}{r^2} - \frac{B}{r}$,where $A$ and $B$ are positive constants and $r$ is the distance of the particle from the centre of the field. For stable equilibrium,the distance of the particle is:

$A$ particle moving along the $X$-axis has potential energy $U = 2 - 20x + 5x^2 \text{ J}$. The particle is released at $x = -3 \text{ m}$. What is the maximum value of $x$ reached by the particle (in $\text{ m}$)? ($x$ is in meters and $U$ is in joules)

The potential energy of a body of mass $m$ is given by $U = ax + by$,where $x$ and $y$ are the position coordinates of the particle. The acceleration of the particle is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module