Thank you for the opportunity to help you with your question!
m_1*v_1+m_2v_2=(m_1+m_2)v_3 The left side of the equation is 3 car momentum added to the 1 car momentum before the collision. The right side of the equation is the momentum of all 4 cars after the collision. Momentum=mv for any object.
The space _ denotes sub 1,2, or 3 to show which object it is.
If the first three trains have mass 3m because each one has mass m. These represent mass number 1. It is a variable since we do not know its actual mass. They have a velocity of 1.5. The second train or mass number 2 is just m because it is identical to the other three train cars. This car has velocity 0 because it is not moving at the beginning of the problem. The equation can now be filled in to solve for the third velocity aka the 4 car train.
(3m)(1.5)+(m)(0)=(3m+m)v_3 I plugged in for what I know.
4.5m+0=4m(v_3) I simplified by putting terms together.
4.5m=4m(v_3) I dropped the 0. Now divide by 4m.
9/8=v_3 This is the final velocity of all the masses combined at the time after the collision when the 4 car train is moving.
How would I solve for this portion of the problem>
The four-car train rolls down the track without losing speed until they collide with and couple to a fifth car that is stationary on the train tracks. If the fifth car is identical to other four, what is the speed of the five-car train?
Use the same formula for before and after the collision
Without retyping and plugging in it looks like this...
36m/8m=5mv Divide by 5m.
36/40=v Reduce the fraction and make it a decimal. Can be done with a calculator 36/40=.9
.9=v This is the velocity of the 5 car train.
Sorry. There was something wrong with your post. There was information missing. But I see it now.
The 4th car originally is moving at 4.2 not 0.
8.7m=4mv Divide by 4m
Use 2.175 instead of 9/8 in the other equation.
4m(2.175)+m(0)=(4m+m)v It is 0 because the fifth car is stationary aka not moving.
8.7m=5mv Divide by 5m.
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