11. The mobile shown below is perfectly balanced. What must be the masses of m.,

Question

11. The mobile shown below is perfectly balanced. What must be the masses of m., ms, and m,? Show your work below. Figure 11-3 3 g my 3 8 an2 12. Recall that the momentum of an object is equal to the product of its mass and velocity. A 1.0 kg object moving at 3.0 m/s collides elastically with a 2.0 kg stationary object; after the collision the 2.0 kg object moves forward (same direction of object 1 before the collision) with a speed of 2.0 m/s. After the collision the 1.0 kg object moves a) forward with a speed of 2.0 m/s b) forward with a speed of 1.0 m/s c) forward with a speed of 3.0 m/s d) backward with a speed of 1.0 m/s e) backward with a speed of 2.0 m/s

Explanation / Answer

(11)The moment of the weights should perfectly balance on either side. So, taking moments about the point of hanging, ( moment = weight * horizontal distance ) and summing for each side and equating for both sides; 3*13+5*m1+2*m2 = 9*m3+6*1+2*3_____________(equation A)

Doing this for lower branches also; 3*m3 = 4*3 or m3 = 4g

also, 1*m2 = 2*m1 or m2 = 2*m1

So putting m2 = 2* m1 and m3 = 4g in equation A, we get m1 = 1g and m2 = 2g

(12)Initial momentum = 1*3 + 2*0 = 3kgm/s = Final momentum = 2*2+1*v

or, 3-4 = 1*v

or, v = -1m/s

so, (d) backward with a speed of 1m/s

please upvote to appreciate

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