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A 1.2-kilogram block and a 1.8-kilogram block are initially at rest on a frictionless, horizontal surface.when a compressed spring between the blocks is released,the 1.8-kilogram block moves to the right at 2.0 meters per second, as shown. what is the speed of the 1.2-kilogram block after the spring is released? *

Respuesta :

Louli
For this problem, we will assume that:
the right direction has a positive sign
the left direction has a negative sign
Note: You can assume the opposite but take care during the substitution

The law of conservation of momentum states that the total momentum of the system remains constant unless an external force acts on this system.

This law can be translated into the following equation:
m1v1 + m2v2 = m1v1' + m2v2'
where:
m1 is mass of first object = 1.2 kg
v1 is the initial velocity of first object = 0 m/sec (starts from rest)
m2 is the mass of the second object = 1.8 kg
v2 is the initial velocity of second object = 0 m/sec (starts from rest)
v1' is the final velocity of the first object that we need to find
v2' is the final velocity of the second object = 2 m/sec

Substitute with the givens in the above equation to get v1' as follows:
m1v1 + m2v2 = m1v1' + m2v2'
(1.2)(0) + (1.8)(0) = (1.2)(v1') + (1.8)(2)
0 =1.2v1'  + 3.6
1.2v1' = -3.6
v1' = -3 m/sec

This means that the final velocity of the first object is 3 m/sec in the left direction

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