Suppose that each of the three masses in the figure below has a mass of 6.40 kg,

Question

Suppose that each of the three masses in the figure below has a mass of 6.40 kg, a radius of 0.0721 m, and are separated by a distance of 8.5 m. If the balls are released from rest, what speed will they have when they collide at the center of the triangle? Ignore gravitational effects from any other objects. Your response differs from the correct answer by more than 10%. Double check your calculations, mu m/s An equilateral triangle 64.0 m on a side has a m_1 = 35.00 kg mass at one corner, a m_2 = 50.00 kg mass at another corner, and a m_3 = 65.00 kg mass at the third corner. Find the magnitude and direction of the net force acting on the 35.00 kg mass. Your response differs from the correct answer by more than 10%. Double check your calculations.degree counterclockwise from the positive x-axis

Explanation / Answer

the potential energy is, U = -Gm1m2/r12

The energy is,

    U1 = -G*(m1m2/r12 + m1m3/r13 + m2m3/r23)

if m1 = m2=m3 = m and r12 = r13 = r23 =r, we get

            U1 = -3G*m2/r2

                  = - 3*6.67x10-11*((6.4/8.5)^2)

                  = -11.34x10^-11J

When they collide, the value of r is reduced to,

             r = 2*0.0721 =0.1442 m

So U2 final = -3*G*(6.4/0.1442)^2 = -3.94x10^-7 J

but, U1 = K2 + U2

so,

    K2 = U1-U2 = -11.34x10^-11 + 3.94x10^-7 = 3.9405x10^-7J

Thus, the velocity is,

           3*(1/2*m*v^2) = 3.9405x10^-7

                         v = 2.026x10-4 m/s = 202.6x10-6 m/s = 202.6 micro.m/s

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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