Answer:
K.E_f = 6.0771 *10^-8 J
F_net = - 8.58 * 10^-8 i N
Explanation:
Given:
- Two sphere masses m_1 & m_2 = 21 kg
- A ball of mass m_3 = 10 kg
- Sphere placed at y = 0.35 m and y = -0.35 m
- Initial distance of ball @ rest from spheres is infinite
- The ball reaches @ ( x , y ) = ( 0.3 , 0 ) m
Find:
(a) its kinetic energy
(b) the net force on it from the spheres, in unit-vector notation?
Solution:
- The conservation of Energy gives us that total energy of the system of 2 spheres and ball is conserved as follows:
(KE_i + 2*U_i) = (KE_f + 2*U_f)
- We know that the ball was initially at infinity and released from rest. We can say that (KE_i + 2*U_i) = 0. KE_i = 0 and U_i = 0. Hence,
K.E_f = 2*U_f
- Where, U_f is the gravitational potential energy of the ball and one sphere. Which can be expressed as:
K.E_f = 2*G*m_3*m_2 / r
- Now we calculate the distance of ball from one of the spheres i.e r. Using Pythagoras Theorem:
r = sqrt ( 0.35^2 + 0.3^2) = 0.460977 m
- Now evaluate the K.E_f:
K.E_f = 2*6.67*10^-11*10*21 / 0.460977
K.E_f = 6.0771 *10^-8 J
- The gravitational force acting on the ball due to two spheres is:
F_1 + F_2 = F_net
- We see that the vertical component of F_1 and F_2 forces acting on ball due to identical spheres with similar distance r is cancelled out. Hence, only the component along the x-axis acts on the ball as a resultant:
-F*cos(Q) + -F*cos(Q) = F_net
F_net = -2*F*cos(Q)
- We can compute this angle Q, using trigonometry as follows:
Q = arctan ( 0.35 / 0.3 ) = 49.4 degrees
- Now compute the required Net force:
F_net = -2*cos(Q)*Gm_1*m_3 / r^2
F_net = -2*cos(49.4)*6.67*10^-11*10*21 / 0.460977^2
F_net = - 8.58 * 10^-8 i ... (towards - x direction)