Question

question 2
suppose m1 (2 kg) and m2 (4 kg) are connected by a compressed spring and are at rest on a frictionless surface. if the spring is allowed to expand and push the masses apart, giving m1 a velocity of v1 = -3 m/s, what will be the final velocity of m2?
a -3 m/s
b +3 m/s
c -6 m/s
d -6 m/s
e 1.5 m/s
f -1.5 m/s

Explanation:

Step1: Apply the law of conservation of momentum

The initial momentum \(P_i = 0\) (since the system is at rest initially). According to the law of conservation of momentum \(P_i=P_f\), where \(P_f = M_1v_1+M_2v_2\). So, \(0 = M_1v_1+M_2v_2\).

Step2: Solve for \(v_2\)

We know \(M_1 = 2\space kg\), \(v_1=- 3\space m/s\) and \(M_2 = 4\space kg\). Substitute these values into the equation \(0 = M_1v_1+M_2v_2\).

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Answer:

E. \(1.5\space m/s\)