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Mathematics 19 Online
OpenStudy (anonymous):

two railway trucks of masses m and 3m move towards each other in opposite direction with speeds 2v and v respectively . these trucks stick together . what is the total speed of trucks after the collision?

OpenStudy (zzr0ck3r):

conservation

OpenStudy (zzr0ck3r):

3mv - 2mv = 4sm solve for s

OpenStudy (zzr0ck3r):

are you doing momentum conservation?

OpenStudy (anonymous):

the correct answer is v/4 but i dont understand how i can get it

OpenStudy (zzr0ck3r):

3mv - 2mv = 4sm so 4sm=mv so 4s=v so s=v/4

OpenStudy (anonymous):

@zzr0ck3r yes it is connservation of momentum

OpenStudy (zzr0ck3r):

understand?

OpenStudy (anonymous):

yes. best response :)

OpenStudy (anonymous):

Let's denote the direction the mass \(m\) is moving in as positive. We know that in every collision momentum is conserved. Let's compute our initial momenta:$$p_0=m(2v)=2mv\\p_1=(3m)(-v)=-3mv$$... so our net momentum is given by \(2mv-3mv=-mv\). After the collision, we're left with one body of combined mass, i.e. \(M=m+3m=4m\). We wish to determine its speed. We find its momentum to be \(Mv'=4mv'\). As momentum is conserved, we know \(-mv=4mv'\). Now solve for \(v'\):$$-mv=4mv'\\-v=4v'\\v'=-\frac14v$$

OpenStudy (anonymous):

i.e. post-collision we will be travelling in the direction of \(3m\) mass at a quarter of its initial speed. :-) this should make intuitive sense.

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