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joemomma
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Homework Statement
Two mine cars of equal mass are connected by a rope which is initially slack. Car A is imparted a velocity of 4m/s with car B initially at rest. When the slack is taken up the rope suffers a tension impact that imparts a velocity to car B and reduces the velocity of car A.
a. If 40% of the kinetic energy of car A is lost during the impact calculate the velocity imparted to car B.
b. Following the initial impact car B overtakes car A and the two are coupled together. Calculate their final common velocity.
Homework Equations
The definition of kinetic energy:
KE = \frac{1}{2}*m*V^{2}
Momentum impulse equation:
m_{1}*v_{1}+\int\sum F = m_{2}*v_{2}
The Attempt at a Solution
part B is easy, you just assume both carts are a system so there is no outside impulse.
m_{B}*0+m_{A}*4=m_{A}*v+m_{B}*v
m_{A} = m_{B}
Since they are both the same mass they each end up going 2 m/s.
Part A is confusing me. The problem states that 40 % of the kinetic energy of cart A is lost during the impact. That means that 60 % of the kinetic energy is remaining. I should be able to use the definition of kinetic energy to solve for the velocity of cart A after the impact by saying:
0.6*KE_{A1}=KE_{A2}
After that I'm not really sure what to do. I presume the change in kinetic energy can be seen as an impulse, where
M*V_{2}-M*V_{1} = \int\sum F
With M being the mass of one cart, V1 being the initial speed of cart A (4 m/s), and V2 being the speed of cart A after the impact. Then that impulse would be applied to cart 2 with an initial speed of 0 m/s.
The problem is that when I take this method I solve for a post-impact speed of cart A being 3.1 m/s and cart B being 0.9 m/s. Obviously this can't be right since the stationary cart is supposed to be overtaking the cart that was originally moving.
I hope that made sense. I'm not really used to this LATEX formatting. If I can elaborate on anything please let me know.