# Check answer for momentum prob

1. Dec 12, 2003

### 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?

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

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

$$v'=3.52 m/s$$

2. Dec 12, 2003

### Staff: Mentor

Note that they meet head on. Direction matters!

3. Dec 12, 2003

### bard

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

i dont understand what u mean by "direction matters"

4. Dec 12, 2003

### Staff: Mentor

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!

5. Dec 12, 2003

### bard

then i would get a negative v

since

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

6. Dec 12, 2003

### Staff: Mentor

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

7. Dec 12, 2003

### bard

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

Last edited: Dec 12, 2003