Consider a system of three particles with masses 0.1 kg, 0.2 kg, and 0.3 kg. The

ID: 1265413 • Letter: C

Question

Consider a system of three particles with masses 0.1 kg, 0.2 kg, and 0.3 kg. The velocity of the center of mass is initially zero.

a) From that information, do you know the position, velocity, or acceleration of the 0.1 kg particle? Explain.

b) A force of 1.5 N is then applied to the particle of mass 0.1 kg. How will the center of mass of the system be affected by the force?

c) Would your answer to (a) change if the force were applied to m2 instead of m1? Explain.

d) If the force is applied for 5 seconds, explain why the total final momentum must be p_tot = 7.5 kg m/s.

e) What is the final velocity of the center of mass?

No, from this infomation , I can't tell about position , velocity or acceleration of the 0.1 kg particle.

here , when the force 1.5 N applied , using second law of motion

1.5 =(.1 + .2 + .3) * a

a = 2.5 m/s^2

the acceleraion of the centre of mass is 2.5 m/s^2

no, the answer to a stays the same

Using second law of motion

Change in momentum = force * time

Change in momentum = 1.5 * (5)

Change in momentum = 7.5 kg.m/s

Now , using

change in momentum = impulse

m*(vf - vi) = 7.5

0.6 *(vf) = 7.5

vf = 12.5 m/s

the final velocity of centre of mass is 12.5 m/s

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