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 particle oscillating along the $x-$axis is given by $U = 20 + (x - 2)^2$,where $U$ is in joules and $x$ is in meters. The total mechanical energy of the particle is $36\,J$. The maximum kinetic energy of the particle is ................ $J$.

What does a body at rest possess?

$A$ body at rest may have

Potential energy is a......

$A$ particle has a total mechanical energy that is small and negative. It is under the influence of a one-dimensional potential $U(x) = \frac{x^4}{4} - \frac{x^2}{2} \, J$,where $x$ is in meters. At time $t = 0 \, s$,it is at $x = -0.5 \, m$. Then,at a later time,it can be found:

Vedclass Test Series

Exam Paper Generator

Online Exam Module