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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?

Sagot :

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.

[tex]p_{0_1}+p_{0_2}=p_{f_1}+p_{f_2}[/tex]

Using the definition of momentum p = mv, we have

[tex]m_1\cdot v_{0_1}+m_2\cdot v_{0_2}=m_1\cdot v_{f_1}+m_2\cdot v_{f_2}[/tex]

Replacing the given information, we have

[tex]1\operatorname{kg}\cdot5m/s+3\operatorname{kg}\cdot2m/s=1\operatorname{kg}\cdot4m/s+3\operatorname{kg}\cdot v_{f_2}[/tex]

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

[tex]\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}[/tex]

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