3 boxcars moving along frictionless track hits a motionless boxcar

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Three boxcars moving at 20 m/s collide with a stationary boxcar on a frictionless track. The conservation of momentum is applied, resulting in the equation 60 = 4v_f, where v_f is the final speed of the coupled boxcars. Since all boxcars have equal mass, the masses cancel out in the equation. The final speed of the four coupled boxcars is calculated to be 15 m/s. This solution confirms the correct application of physics principles to solve the problem.
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Homework Statement


Three boxcars are rolling along a virtually frictionless track at 20m/s when they strike another boxcar sitting motionless on the track. what is the new speed of the 4 coupled bovcars? all the boxcars have the same mass

Homework Equations


3rd or 2nd Newton law

The Attempt at a Solution


m_a(20) + m_b(20) + m_c(20) + m_d(0) = m_a(v_f) + m_b(v_f) + m_c(v_f) + m_d(v_f)
since all of the masses are equal i cross all of them out.
60 = 4v_f
v_f = 15m/s
 
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chaotiiic said:

Homework Statement


Three boxcars are rolling along a virtually frictionless track at 20m/s when they strike another boxcar sitting motionless on the track. what is the new speed of the 4 coupled bovcars? all the boxcars have the same mass


Homework Equations


3rd or 2nd Newton law


The Attempt at a Solution


m_a(20) + m_b(20) + m_c(20) + m_d(0) = m_a(v_f) + m_b(v_f) + m_c(v_f) + m_d(v_f)
since all of the masses are equal i cross all of them out.
60 = 4v_f
v_f = 15m/s

That is the correct answer to the problem you described.
 
PeterO said:
That is the correct answer to the problem you described.
thankyou.
 

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