Two particles, 1 and 2, with masses m1-8x1025 kg and m2 = 2× 10-25 kg, are const

Question

Two particles, 1 and 2, with masses m1-8x1025 kg and m2 = 2× 10-25 kg, are constrained to move in the x-y plane. They interact by a repulsive central force, F kri2 where ri2 is the distance between the particles. The diagram shows the situation. The force on particle 2 is given by where is the unit vector in the direction of ri2, with the x-y coordinates of the particles, Xi i,i-1, 2. Note that a unit vector has length ; its components are numbers and have no units Let k-0.8x1025 N m2. Assume the initial components of positions (in meters) and velocities (in m/s) of the particles as given in the table: 500000.00 000000 30000 .1 20000 | Vir, | 000000 000000060000 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. [Ipt Calculate the initial distance between the particles, r12. Correct, computer gets:2.87924@+00m 21. [lpt) Calculate the x component of the unit vector 12 Correet, computer gets1-9.37749e-01 22. [lpt) Calculate the y component of 2 Correct, computer gets:-3.47314e-01 23. lpt] Calculate the x component of F2. Answe 24. [lpt Calculate the y component of F2. Answer:

Explanation / Answer

Given x1=1.5 y1 =1    x2=-1.2    y2=0 Vx1=0 Vy1=0.3 Vx2=0    Vy2=-0.6

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

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

r12 =(x2-x1)i +(y2-y1)j =(-1.2-1.5)i+(0-1)j =(-2.7)i+(-1)j

lr12l = SQRT((x2-x1)^2+(y2-y1)^2) =SQRT((-2.7)^2+(-1)^2)=2.87923601

unit vector in direction of r12 = r12/lr12l =(-2.7i-j)/2.87923601 =(-0.93775)i+(-0.34731)j

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

   = ((0.8*10^-25)/(2.87923601)^2)(-0.93775i-0.34731j)

   =(9.650180939*10^-27)(-0.93775i-0.34731j)

   F =(-9.049457176*10^-27)i+(-3.351604342*10^-27)j

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

F2= (-9.049457176*10^-27)i+(-3.351604342*10^-27)j

23.) x component of F2 = -9.049457176*10^-27

24) y component of F2 = -3.351604342*10^-27

25) F 1 =-F2 = (9.049457176*10^-27)i+(3.351604342*10^-27)j

   X component of F1 =9.049457176*10^-27

26) a1 =F1/m =((9.049457176*10^-27)i + (3.351604342*10^-27)j)/(2*10^-25)

   a1 = (0.04524728588)i+(0.01675802171)j

   initial v1=vx1 i +vy1 j=0i+0.3j

v1 new = initial v1 +a1t

given t=0.1 sec

V1 new = (0i+0.3j)+(0.04524728588i+0.01675802171j)(0.1)

   V1 new = (0.004524728588i)+(0.30167580217)j

X component of V1 new =0.004524728588

27) Y component of V1 new =0.30167580217

28) V2 new = v2 old + a2t

   a2=F2/m=-F1/m=-a1

a2=(-0.04524728588)i +(-0.01675802171)j

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

   V2 old =0i+(-0.6)j

V2 new = V 2 old + a2 t

   V2 new = (0i -0.6j)+((-0.04524728588)i+(-0.01675802171)j)0.1

V2 new = (-0.004524728588)i+(-0.60167580171)j

X component of V2 new =-0.0045247258588

29) Y component of V2 new =-0.60167580171

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

r1 old =1.5i+1 j

   r1 new = (1.5i+1j)+((0.0045247i+0.3016758j)0.1 +0.5(0.045247i+0.016758j)0.01)

   = (1.5i+j)+(0.000678705i+0.03025137j)

   r1 new = 1.500678705i+1.03025137j

X component of r1 new = 1.5006787031)

31) y component of r1 new =1.03025134

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

   r2 old =1.2i+0j

r2 new = (1.2i+0j)+((-0.0045247i-0.601675j)0.1+(0.5(-0.045247i-0.016758j)0.01))

   r2 new =(1.2i+0j) +(-0.000678705-0.06025123)

   r2 new = 1.199321295i-0.06025123j

X component of r2 new =1.199321295

33) Y component of r2 new =-0.06025123

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