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mmA 1.0 kg cart moving right at 5.0 on a frictionless track collides with a 3.0 kg cart moving right at 2.0The lighter cart has a final speed of 4.0 to the right.SmWhat is the final speed of the 3.0 kg cart?

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To answer this, we have to use the conservation of momentum law, which indicates that the sum of initial momentums is equal to the sum of final momentums after a collision.


p_(0_1)+p_(0_2)=p_(f_1)+p_(f_2)

Using the definition of momentum p = mv, we have


m_1\cdot v_(0_1)+m_2\cdot v_(0_2)=m_1\cdot v_(f_1)+m_2\cdot v_(f_2)

Replacing the given information, we have


1\operatorname{kg}\cdot5m/s+3\operatorname{kg}\cdot2m/s=1\operatorname{kg}\cdot4m/s+3\operatorname{kg}\cdot v_(f_2)

Then, we solve for the final velocity of the heavier car


\begin{gathered} 5\operatorname{kg}\cdot m/s+6\operatorname{kg}\cdot m/s=4\operatorname{kg}\cdot m/s+3\operatorname{kg}\cdot v_(f_2) \\ 11\operatorname{kg}\cdot m/s-4\operatorname{kg}\cdot m/s=3\operatorname{kg}\cdot v_(f_2) \\ 7\operatorname{kg}\cdot m/s=3\operatorname{kg}\cdot v_(f_2) \\ v_(f_2)=\frac{7\operatorname{kg}\cdot m/s}{3\operatorname{kg}} \\ v_(f_2)\approx2.3m/s \end{gathered}

Hence, the final speed of the 3.0 kg cart is 2.3 m/s, approximately.

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