Question

2)Two masses M move at speed V, one to the east and one to the west. What is the total energy of the system?a.Now consider the setup as viewed from a frame moving to the west at speed u. Find the energy of each mass in this frame. Is the total energy larger or smaller than the total energy in the lab frame?

Answer 1

Answer:

Part A: E = MV²

Part B: E1/3 = energy of the first car relative to the moving frame = 1/2×M×(V + u)²,

E2/3 = energy of the first car relative to the moving frame= 1/2×M×(u – V)²

Part C: Et/3 = Total Energy of the System = M(V² + u²)

Explanation:

Part A

Taking the east direction as positive,

Vehicle 1: mass = M, velocity = V

Vehicle 2: mass = M, velocity = – V (west)

In the first case where both vehicles are viewed from a stationary frame, the total energy of the system is

E = 1/2×MV² + 1/2×M(–V)²

E = 1/2×MV² + 1/2×MV²

E = MV²

Part B

Now Case 2: moving frame

Vehicle 1: mass = M, velocity = V

Vehicle 2: mass = M, velocity = – V (west)

Vehicle 3: velocity = – u (west)

For this case we need to find the velocities of vehicles 1 and 2 relative to vehicle 3

Let V1/3 be the velocity of vehicle 1 relative to 3 and

V2/3 be the velocity of vehicle 2 relative to 3

V1/e = velocity of vehicle 1 relative to the earth = V

V2/e = velocity of vehicle 2 relative to the earth = – V

V3/e = velocity of vehicle 2 relative to the earth = – u

(still choosing the east direction as positive)

Ve/3 = velocity of earth relative to vehicle 3 = – V3/e = – (–u) = u

So taking the earth as a stationary frame of reference,

V1/3 = V1/e + Ve/3

V1/3 = V + u

V2/3 = V2/e + Ve/3

V2/3 = –V + u

V2/3 = u – V

E1/3 = 1/2×M×V1/3²

E1/3 = 1/2×M×(V + u)²

E1/3 = 1/2×M×V1/3²

E2/3 = 1/2×M×(u – V)²

Part C

Let

Et/3 = Total energy relative to the moving frame = 1/2×M×(V + u)² + 1/2×M×(u – V)²

Et/3 = 1/2×M×(V² + 2Vu +u² + u² – 2Vu + V²)

Et/3 = 1/2×M×(2V² + 2u²)

Et/3 = 1/2×2×M×(V² +u²)

Et/3 = M(V² + u²)

The total energy of the system increases by Mu²