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## Momentum Skill Based Problems File

```Check answers to 65
J= change in P
J=Pfx-Pox
J=-.145*38*cos36-.145*38*cos36
J=2(.145*38*cos36)
J=8.9kgm/s```
```Skill Building Problem #65:
I found the initial momentum =(.145)(38)=5.51 kg*m/s, which is equal to the final momentum. I then found initial py = final py = 5.51(sin36), initial px = 5.51cos36, and final px = -5.51cos36.  So impulse in the 7 direction is 0, and impulse in the x direction is -2(5.51cos36).  I got -8.92 Ns along the x axis as the total impulse, but the key says it should be 61.5 Ns.  I can&#x27;t figure out what I&#x27;m doing wrong.```
`Just checking on my previous comment.`
`Christopher and Savannah - sorry for the delay in the answer - the question got lost!  You are correct. The magnitude of the momentum with 2 significant figures is 8.9 kg m/s or Ns.  Thanks for the correction. John`
```#40, why are the velocities both positive? shouldn&amp;amp;amp;amp;#x27;t one be negative and one positive?

also we got 32 m/s for the ball, not 115 m/s```
```Ben, the problem, when solved with the relative velocity equation yields -114 m/s, which means the ball reverses direction and takes off at roughly 230 mph!  Not very realistic. When using the conservation of momentum equation, you do get 32 m/s, which means that the ball goes through the bat.
`Furthermore, the relative velocity equation for elastic collisions was derived from the conservation of KE and the conservation of momentum equations solved simultaneously. At one point in the derivation, we divided by (v2 - v2'). For this problem the bat is v2 and v2' and since its velocity doesn't change, (v2-v2') = 0.  Can't do that!  So that equation doesn't work for this problem.`
`Ben,the problem is fixed.  Please check it out.  John`