Two particles, 1 and 2, with masses m1- 8x10-25 kg and m2 8x10-25 kg, are constr

Question

Two particles, 1 and 2, with masses m1- 8x10-25 kg and m2 8x10-25 kg, are constrained to move in the x-y plane. They interact by a repulsive central force, FEk/rn2 where 12 is the distance between the particles. The diagram shows the situation T12 The force on particle 2 is given by where is the unit vector in the direction of r12 r12 = (X2 -x1)^ + U2 with the x-y coordinates of the particle:s Xi, yi, i 1, 2. Note that a unit vector has length 1; its components are numbers and have no units Let k= 0.8×10-25 N·m2. Assume the initial components of positions (in meters) and velocities (in m/s) of the particles as given in the table old 0.00000 yold o.500001.00000 x2old -1.90000ld0.00000 0.00000y0.30000 xyold1.80000 x Now we will use the ideas of molecular dynamics developed in class to determine how the particles move under the influence of this force. Use 6 sig. figs. in your answers! 20. [1pt] Calculate the initial distance between the particles, r12. Answer: 21. [1pt] Calculate the x component of the unit vector 2. Answer 22. [1pt] Calculate the y component of f12 Answer Submit All Answers Submit All Answers Submit All Answers

Explanation / Answer

Given x1=1.8 y1 =0.5    x2=-1.9    y2=0 Vx1=0 Vy1=1.0 Vx2=0    Vy2=-0.3

k=0.8*10^-25 Nm^2 ,m =8*10^-25 kg

F= (k/(lr12l)^2). (r12/lr12l)

21)r12 =(x2-x1)i +(y2-y1)j =(-1.9-1.8)i+(0-0.5)j =(-3.7)i+(-0.5)j

   lr12l = SQRT((x2-x1)^2+(y2-y1)^2) =SQRT((-3.7)^2+(-0.5)^2)=3.7336

unit vector in direction of r12 = r12/lr12l =(-3.7i-0.5j)/3.7336 =(-0.9910)i+(-0.1339)j

21)X component of unit vector =-0.9910   

22) Y component of unit vector =-0.1339

Force vector F = (K/(r12)^2). (r12/lr12l)

   = ((0.8*10^-25)/(3.7336)^2)(-0.9910i-0.1339j)

   =(2.1427*10^-26)(-0.9910i-0.1339j)

   F =(-2.1234*10^-26)i+(-0.2843*10^-26)j

Force acting on F1 is equal opposite in direction to the force acting on F2 ,F1=-F2

F2= (-2.1234*10^-26)i+(-0.2843*10^-26)j

23.) x component of F2 = -2.1234*10^-26

24) y component of F2 = -0.2843*10^-26

25) F 1 =-F2 = (2.1234*10^-26)i+(0.2843*10^-26)j

   X component of F1 =2.1234*10^-26

26) a1 =F1/m =((2.1234*10^-26)i + (0.2843*10^-26)j)/(8*10^-25)

   a1 = (0.02654)i+(0.003553)j

   initial v1=vx1 i +vy1 j=0i+1j

v1 new = initial v1 +a1t

given t=0.1 sec

V1 new = (0i+1j)+(0.02654i+0.003553j)(0.1)

   V1 new = (0.02654i)+(1.003553)j

X component of V1 new =0.02654

27) Y component of V1 new =1.003553

28) V2 new = v2 old + a2t

   a2=F2/m=-F1/m=-a1

a2=(-0.02654)i +(-0.003553)j

V2 old =(Vx2)i+(Vy2)j

   V2 old =0i+(-0.3)j

V2 new = V 2 old + a2 t

   V2 new = (0i -0.3j)+((-0.02654)i+(-0.003553)j)0.1

V2 new = (-0.02654)i+(-0.303553)j

X component of V2 new =-0.02654

29) Y component of V2 new =-0.303553

30) r1 new = r1 old + ((V1 new) t+0.5 a1(t^2))

r1 old =1.8i+0.5 j

   r1 new = (1.8i+0.5j)+((0.02654i+1.003553j)0.1 +0.5(0.02654i+0.003553j)0.01)

   = (1.8i+0.5j)+(0.002787i+1.003570j)

   r1 new = 1.802787i+1.503570j

X component of r1 new = 1.802787

31) y component of r1 new =1.503570

32) r2 new = r2 old +(( V2 new)t+(0.5(a2)t^2))

   r2 old =-1.9+0j

r2 new = (-1.9i+0j)+((-0.02654i-0.303553)0.1+(0.5(-0.02654i-0.003553j)0.01))

   r2 new =(-1.9i+0j) +(-0.002787i-0.030373j)

   r2 new = -1.902787i-0.030373j

X component of r2 new =-1.902787

33) Y component of r2 new =-0.030373

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.