A red cart moves toward the right at speed v on a frictionless track while a blue cart moves toward it at speed 2v. The mass of the blue cart is m while the mass of the red card is 2m. Before the carts collide: (a) What is the momentum of the two-cart system, in terms of m and v?

Taking the right direction as positive, we calculate the initial momentum [tex]p_i[/tex] as the sum of both the red cart initial momentum [tex]p_R=m_Rv_R=(2m)(v)[/tex] and blue cart initial momentum [tex]p_B=m_Bv_B=(m)(2v)[/tex], where since the blue cart moves towards it at twice the velocity we understand that it also goes towards the right direction to meet the red cart for the collision (if it goes towards the left its velocity would be -2v, but that would be moving against the red cart).

Thus we have:

[tex]p_i=2mv+m2v=4mv[/tex]

Muscardinus Muscardinus · Answer 2 · 2019-12-17T08:53:43+01:00

Answer:

Net momentum, [tex]p=4mv[/tex]

Explanation:

Let m is the mass of blue cart and 2m is the mass of red cart. v is the speed of red cart and 2v is the speed of blue cart. Net momentum is equal to the sum of momentum of blue and red cart.

Net momentum = momentum of red cart + momentum of blue cart

[tex]p=p_r+p_b[/tex]

[tex]p=2m\times v+m\times 2v[/tex]

[tex]p=4mv[/tex]

So, the momentum of the two-cart system is 4 mv. Hence, this is the required solution.