An 8.00-kg point mass and a 14.0-kg point mass are held in place 50.0 cm apart.

Question

An 8.00-kg point mass and a 14.0-kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point between the two masses 15.0 cm from the 8.00-kg mass along the line connecting the two fixed masses. Find the magnitude of the initial acceleration of the particle. Express your answer with the appropriate units. What is the direction of the initial acceleration of the particle? Acceleration of the particle is toward the 8.00-kg mass. Acceleration of the particle is toward the 14.0-kg mass.

Explanation / Answer

A)
Both object will attract it

Let 14 Kg mass is placed right of 8 Kg mass

In this F due to 14 kg will be towards right and hence positive

a = Fnet / m
= (F due to 14 Kg - F due to 8 Kg) /m
=G*m* (M1/R1^2 - M2/R2^2) /m
=G* (M1/R1^2 - M2/R2^2)

Here, M1 = 14 Kg
R1= 50-15 = 35 cm = 0.35 m
M2 = 8 Kg
R2 = 15 cm = 0.15 m

a =G* (M1/R1^2 - M2/R2^2)
= (6.673*10^-11)*(14/0.35^2 - 8/0.15^2)
= (6.673*10^-11)*(114.3 - 355.6)
= -1.6*10^-8 m/s^2
Answer: 1.6*10^-8 m/s^2

b)
begative sign shows that it is directed toward 8 Kg
Answer: 1st option

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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