##### Physics: Conservation of momentum

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Three identical train cars, coupled together, are rolling east at 1.5

m/s . A fourth car traveling east at 4.2  catches up with the three and couples to make a four-car train.

what is the speed of the 4 car train ?

Oct 25th, 2015

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.

Oct 25th, 2015

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?

Oct 25th, 2015

Use the same formula for before and after the collision

Without retyping and plugging in it looks like this...

4m(9/8)+m(0)=(4m+m)v

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.

Oct 25th, 2015

I am doing my homework online and it says that this answer is incorrect

Oct 25th, 2015

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.

3m(1.5)+m(4.2)=(3m+m)v

4.5m+4.2m=4mv

8.7m=4mv Divide by 4m

8.7m/4m=v

2.175=v

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.

8.7m/5m=v

1.74=v

Oct 25th, 2015

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Oct 25th, 2015
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Oct 25th, 2015
Oct 19th, 2017
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