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marissa [1.9K]
3 years ago
11

Is it possible for the momentum of a system consisting of two carts on a low-friction track to be zero even if both carts are mo

ving?
Physics
1 answer:
choli [55]3 years ago
4 0

Answer:

Yes, if the two carts are moving into opposite directions

Explanation:

The total momentum of the system of two carts is given by:

p=m_1 v_1 + m_2 v_2

where

m1, m2 are the masses of the two carts

v1, v2 are the velocities of the two carts

Let's remind that v (the velocity) is a vector, so its sign depends on the direction in which the cart is moving.

We want to know if it is possible that the total momentum of the system can be zero, so it must be:

p=0\\m_1 v_1 + m_2 v_2 = 0\\m_1 v_1 = -m_2 v_2

From this equation, we see that this condition can only occur if v1 and v2 have opposite signs. Opposite signs mean opposite directions: therefore, the total momentum can be zero if the two carts are moving into opposite directions.

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Answer:

Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t

Explanation:

Given that m= 1 slug and given that spring stretches by 2 feet so we can find the spring constant K

mg=k x

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K=16

And also give that damping force is 8 times the velocity so damping constant C=8.

We know that equation for spring mass system

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Now by putting the values

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To find the constant A and B we have to compare this equation with equation 1.

Now find y' and y" (by differentiate with respect to t)

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1 (-16A cos 4t-16B sin 4t)+8( -4A sin 4t+4B cos 4t)+16(A cos 4t+ B sin 4t)=6sin4 t

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y=-3/16 cos 4t

Now to find the CF  of differential equation 1

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Homogeneous version of above equation

m^2+8m+16=0

So CF =(C_1+tC_2)e^{-2t}

So the general equation

Y=(C_1+tC_2)e^{-2t}-3/16 cos 4t

Given that t=0 Y=0 So

C_1=\dfrac{3}{16}

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C_2 =\dfrac{3}{8}

Y=(\dfrac{3}{16}+t \dfrac{3}{8})e^{-2t}-\dfrac{3}{16}cos 4t

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Answer:

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