Is This Calculation of Post-Collision Speed Correct?

[bard]

A 130-KG tackler moving at 2.5 m/s meets head on(and tackles) a 90 k-g halfback moving at 5.0 m/s. What will be their mutual speed immediatley after the collison?

someone please check my work

$$m_{1}v_{1}+m_{2}v_{2}=v'(m_{1}+m_{2})$$

so $$v'=\frac{325+450}{220}$$

$$v'=3.52 m/s$$

 

[Doc Al]

Note that they meet head on. Direction matters!

 

[bard]

Originally posted by Doc Al
Note that they meet head on. Direction matters!

so the $$m_{2}v_{2)$$ becomes negative?

i don't understand what u mean by "direction matters"

 

[Doc Al]

Originally posted by bard
so the $$m_{2}v_{2}$$ becomes negative?

i don't understand what u mean by "direction matters"

Yes, $$m_{2}v_{2}$$ would be negative. If the two collide going the same direction (what you had originally plugged in) you get a totally different answer than if they collide going opposite directions (like in this problem). Momentum is a vector.

In this problem, the motion is along a straight line. So no angles are involved. But the sign sure does matter!

 

[bard]

then i would get a negative v

since

v'=325-450/40=-3.125m/s

 

[Doc Al]

You divided by the wrong number, but yes the answer will be negative. And what does that mean?

 

[bard]

yah i got the answer as -.57m/s. thnx