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Momentum combination

by ProBasket
Tags: momentum
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ProBasket
#1
Apr1-05, 06:23 PM
P: 141
1.)Let us assume that the blocks are in motion, and the x-components of their momenta at a certain moment are [tex]p_1[/tex] and [tex]p_2[/tex], respectively. (there's a pic attached)

Find the x-component of the velocity of the center of mass at that moment.

Express your answer in terms of [tex]m_1,m_2,p_1, p_2[/tex]


my answer is [tex]v_{cm} = \frac{m_1*p_1+m_2*p_2}{m_1+m_2}[/tex] but it's incorrect and i dont understand why.

i think p stands for point particle.

also, i have one more quick quesiton:

2.)Suppose that [tex]v_{cm} = 0 [/tex]. Therefore, the following must be true
A.) [tex]p_1 = p_2[/tex]
B.) [tex] v_1 = v_2[/tex]
c.) m1 = m2
d.) none of the above
for this one, i think that if it's equal to zero, both masses should be equal, is this correct?
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StatusX
#2
Apr1-05, 06:58 PM
HW Helper
P: 2,567
As a general rule, the total momentum of a system is the vector sum of the momenta of all the particles. This is equal to M_total v_cm.

In this case, the position of the center of mass is (m_1 x_1 + m_2 x_2) /(m_1 + m_2). So the velocity is it's change in time, which is (m_1 v_1 + m_2 v_2) /(m_1 + m_2) = (p_1 + p_2)/(m_1+m_2).

For the MC question, remember you can pick the velocity of the center of mass to be anything you want by adding some velocity to both masses. So you can always make it 0 with the right choice of this velocity. Based on what I said in the first paragraph, what does this tell you?


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