2. Two objects collide on a level, frictionless table. The mass of object A is 5

Question

2. Two objects collide on a level, frictionless table. The mass of object A is 5.0 kg; the mass of object B is 3.0 kg. The objects stick together after the collision. The initial velocity of object A and the final velocity of both objects are shown. a. In the space provided, draw separate arrows for object A and for object B representing the direction of the change in momentum vector of the object. Is the magnitude of the change in momentum of object A greater than, less than, or equal to that of object B? Explain your reasoning. b. System C is the system of both objects A and B combined. How does the momentum of system C before the collision compare to the momentum of system C after the collision? Discuss both magnitude and direction. Construct and label a vector showing the momentum of system C at an instant before the collision. Show your work clearly. c. Construct and label a vector showing the initial velocity of object B. Show your work clearly.

Explanation / Answer

mA = 5 kg

mB = 3 kg

From the graphs, VAi = 0.9 i - 0.6 j and VAf = VBf = Vf = 0.3i - 0.6j

From conversation of momentum in inelastic collision,

mA VAi + mB VBi = (mA + mB) Vf

5(0.9 i - 0.6 j) + 3 VBi = 8 (0.3i - 0.6j)

Hence VBi = -0.7i - 0.6j

Change in momentum of A, $PA = mA( Vf - VAi ) = 5[(0.3i - 0.6j) - (0.9 i - 0.6 j)]

Hence $PA = - 3i

Change in momentum of B, $PB = mB( Vf - VBi ) = 3[(0.3i - 0.6j) - (-0.7i - 0.6j)]

Hence $PB = 3i

Magnitude of change in momentum of A is equal to that of B .

System C

System C is both A and B together. Hence PCi = mA VAi + mB VBi = 2.4 i - 4.8j

NOTE : To draw the graphs, divide each answer by the scale of the respective graph to get the number of squares it occupies.

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