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Radda [10]
2 years ago
3

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?
Physics
2 answers:
ki77a [65]2 years ago
5 0

Answer:

p_i=4mv

Explanation:

Taking the right direction as positive, we calculate the initial momentum p_i as the sum of both the red cart initial momentum p_R=m_Rv_R=(2m)(v) and blue cart initial momentum p_B=m_Bv_B=(m)(2v), where since the blue cart moves <u><em>towards</em></u> 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 <em>against</em> the red cart).

Thus we have:

p_i=2mv+m2v=4mv

Airida [17]2 years ago
5 0

Answer:

Net momentum, p=4mv    

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

p=p_r+p_b

p=2m\times v+m\times 2v        

p=4mv    

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

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