Railroad diesel engine homework

[Imparcticle]

A railroad diesel engine weighs four times as much as a freightcar. If the diesel engine coasts at 5km/h into a freightcar that is initially at rest, how fast do the two coast after they couple together?

Given:
m1= 4m2
m2
v1=5km/h
v2=0

unknown : v3

formula:
m1v1+ m2v2=m3v3 (i.e., Total momentum before=total momentum after)

work:

m1v1+ m2v2=m3v3 --> (4 x m2)(5km/h) + (m2)(0)=m3v3
(20km/h)m2=m3v3
answer --> 20km/h=v3

 

[Leong]

sorry to say that it is incorrect. m3 is the total mass after they have coupled together so m3= 4m2 + m2 = 5m2. cancelling m2 at both sides of the equations you finally get v3 = 4 km/h.

 

[Imparcticle]

This really must be wrong:

6.) A comic-strip hero Superhuman meets an asteroid in outer space, and hurls it at 800m/s, as fast as a bullet. The asteroid is a thousand times more massive than Superman. In the strip, Superman is seen at rest after the throw. Taking physics into account, what would be his recoil velocity?



Given unknown formula
M1=1000 M2 v3 m1v1+m2v2=m3v3 --> [( m1v1+m2v2)/m3]=v3
V3=800m/s M2
M3 = (1000M2 + M2) v2
V1= 800m/s

Work: [( m1v1+m2v2)/m3]=v3= {[(1000M2)(800m/s)]+[(M2)(v2)]/[(1000M2 +M2)]}=
{[(M2)(800000 kg * m/s) + [(M2)(v2)]/[(1001 M2]}=
{[(800000 kg * m/s) + [(M2)(v2)]/ 1001}=
799 kg * m/s + [(M2 kg * v2)/1001)]
? I don't get this.

 

[Imparcticle]

m3 is the total mass after they have coupled together so m3= 4m2 + m2 = 5m2.

how do i know that its the mass after they have coupled together?

 

[apchemstudent]

This really must be wrong:

6.) A comic-strip hero Superhuman meets an asteroid in outer space, and hurls it at 800m/s, as fast as a bullet. The asteroid is a thousand times more massive than Superman. In the strip, Superman is seen at rest after the throw. Taking physics into account, what would be his recoil velocity?



Given unknown formula
M1=1000 M2 v3 m1v1+m2v2=m3v3 --> [( m1v1+m2v2)/m3]=v3
V3=800m/s M2
M3 = (1000M2 + M2) v2
V1= 800m/s

Work: [( m1v1+m2v2)/m3]=v3= {[(1000M2)(800m/s)]+[(M2)(v2)]/[(1000M2 +M2)]}=
{[(M2)(800000 kg * m/s) + [(M2)(v2)]/[(1001 M2]}=
{[(800000 kg * m/s) + [(M2)(v2)]/ 1001}=
799 kg * m/s + [(M2 kg * v2)/1001)]
? I don't get this.

You're approaching this question in a wrong way. You are making it more difficult than it really is.

Conservation of momemtum. The system had not momentum at the beginning, which means m1v1+m2v2 = 0.

I really suggest you read your textbook more.

 

[apchemstudent]

how do i know that its the mass after they have coupled together?

As far as this question goes, why do you need to know the mass?

 

[Leong]

Given:
m1= 4m2
m2
v1=5km/h
v2=0

m3=m1+m2
=4m2+m2
=5m2

 

[Imparcticle]

apchemstudent: why? Because of this: m1v1+ m2v2=m3v3

If you're wondering why I asked the question, I was wondering how the quoted person knew "m3 is the total mass after they have coupled together " and not something else.

You're approaching this question in a wrong way. You are making it more difficult than it really is.

Conservation of momemtum. The system had not momentum at the beginning, which means m1v1+m2v2 = 0.

I really suggest you read your textbook more.

I did read,but the book isn't good.

never mind on the question. I missed the last part of the question. I've fixed it. thanks anyway.

 

[Imparcticle]

m3=m1+m2
=4m2+m2
=5m2

thank you very much for your help. :rofl: