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A proton follows the path shown in (Figure 1). Its initial speed is v_{0} = 2.4×10^{6 }m/s .

What is the proton’s speed as it passes through point P?

*you take m*is the weight of the particle

*the velocity of*the speed of the particle , and

*it is*is the diameter of the circle, magnitude of the angular momentum is calculated in terms of the formula

**. A conservation theory of the angular force implies that angular momentum remains constant throughout the circular path.**

*L=mvr*
Apply Energy consrvation,K2 + U2 = K1 + U10.5*m*v^2 + k*q1*q2/r2 = 0.5*m*vo^2 +k*q1*q2/r1

0.5*m*v^2 = 0.5*m*vo^2 + k*q1*q2*(1/r1 – 1/r2)v^2 = vo^2 + (2*k*q1*q2/m)(1/r1 – 1/r2)= (2.4*10^6)^2 +(-2*9*10^9*10*10^-9*1.6*10^-19/1.67*10^-27)*(1/3-1/4)*10^-3

= 5.76*10^12 – 1.43*10^12= 4.33*10^12v = 2.08*10^6 m/s