Two balls of mass 0.5 kg and 0.8 kg are connected by a low-mass spring. This device is thrown through the air with low speed, so air resistance is negligible. The motion is complicated: the balls whirl around each other, and at the same time the system vibrates, with continually changing stretch of the spring. At a particular instant, the 0.5 kg ball has a velocity m/s and the 0.8 kg ball has a velocity m/s.

(a) At this Instant, what is the total momentum of the device?
(b) What is the net gravitational (vector) force exerted by the Earth on the device?
(c) At a time 0.06 seconds later, what is the total momentum of the device?

a) P_total = (4.1 i -1.7 j + 5k) kg m/s

b)  F_g =  12.753 N

c)  P_total = (4.1 i -2.46518 j + 5k) kg m/s

Explanation:

Given:

- The mass of ball 1, m_1 = 0.5 kg

- The mass of the ball 2, m_2 = 0.8 kg

- The velocity of ball 1, V_1 = (5 i -5 j + 2k)

- The velocity of ball 2, V_2 = (2 i +1 j + 5k)

Find:

(a) At this Instant, what is the total momentum of the device?

(b) What is the net gravitational (vector) force exerted by the Earth on the device?

(c) At a time 0.06 seconds later, what is the total momentum of the device?

Solution:

- We will take  unit vectors i , j , k in the directions horizontal, vertical, and out of the page respectively.

- The total momentum of the system is the sum of momentum of individual objects in a system:

                                      P_total = P_1 + P_2

- Where, P_1 : Momentum for ball 1.  P_2 : Momentum for ball 2.

- The momentum of an object is the scalar multiple of the velocity vector and its mass. Hence, we will compute the total momentum as follows:

                                P_total = m_1*V_1 + m_2*V_2

                                P_total = 0.5*(5 i -5 j + 2k) + 0.8*(2 i +1 j + 5k)

Hence, the total momentum is:

                                P_total = (4.1 i -1.7 j + 5k) kgm/s

- The net gravitational Force exerted by the earth on the device is due to the weight of each mass as follows:

                                       F_g = W_1 + W_2

- Where, W_1 : Weight for ball 1.  W_2 : Weight for ball 2.

- Hence, the net gravitational force is as follows:

                                       F_g = m_1*g + m_2*g = g*(m_1 + m_2)

                                        F_g =  9.81*(0.8 + 0.5)

                                        F_g =  12.753 N

- The momentum of the system after a certain time under the experience of gravitational force will affect the initial velocity of the balls in the system. So to calculate the new velocities of the ball. we will apply kinematic equation of motion on j vector of the both balls, in which gravitational acceleration acts.

-                                     V'_1 = (5 i - (5 + g*t) j + 2k)

                                     V'_2 = (2 i +(1 - g*t) j + 5k)  

- Where, V'_1 and V'_2 are new velocities of the ball. Hence, we compute:

                                     V'_1 = (5 i - (5 + 9.81*0.06) j + 2k)

                                     V'_2 = (2 i +(1 - 9.81*0.06) j + 5k)  

- Hence, two velocities are:

                                     V'_1 = (5 i - 5.5886 j + 2k)

                                     V'_2 = (2 i + 0.4114 j + 5k)  

- The momentum of an object is the scalar multiple of the velocity vector and its mass. Hence, we will compute the total momentum as follows:

                                P_total = m_1*V'_1 + m_2*V'_2                

                 P_total = 0.5*(5 i -5.5886 j + 2k) + 0.8*(2 i +0.4114 j + 5k)  

Hence, the total momentum is:

                               P_total = (4.1 i -2.46518 j + 5k) kgm/s

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete

graph question answered .. you think better when you are relaxed and happy