Question

Two 21 kg spheres are fixed in place on a y axis, one at y = 0.35 m and the other at y = −0.35 m. A 10 kg ball is then released from rest at a point on the x axis that is at a great distance (effectively infinite) from the spheres. The only forces acting on the ball are the gravitational forces from the spheres.?

Answer 1

Answer:

K.E_f = 6.0771 *10^-8 J    

F_net = - 8.58 * 10^-8 i N

Step-by-step 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)

Answer 2

Answer:

The question has some details missing. here is the part remaining ; The only forces acting on the ball are the gravitational forces from the spheres.

(a) When the ball reaches the (x, y) point (0.30 m, 0), what is its kinetic energy?

(b) When the ball reaches the (x, y) point (0.30 m, 0), what is the net force acting on it from the spheres, in unit-vector notation? (Express your answer in vector form.

Step-by-step explanation:

The detailed steps and appropriate substitution is as shown in the attachment.