# If the 60 kg player is running forward at 7.0 m/s when she makes contact with the dummy, what is the player’s velocity at the end of the 1.20 s impact? Please can somebody tell me I am doing...

**If the 60 kg player is running forward at 7.0 m/s when she makes contact with the dummy, what is the player’s velocity at the end of the 1.20 s impact? **

**Please can somebody tell me I am doing it right!!**

Impulse can be calculated using either equation:

**Fnet (difference Time)=(difference P(momentum)) or **

**Fnet ( difference Time)=m x (difference velocity)**

**If mass m is constant, then the only quantity changing on the right-hand side of the equation is V(velocity). So the equation becomes:**

**Fnet (difference Time)=m x (difference Velocity)**

**Given I= -360Ns , m=60kg, V initial=7.0 m/s**

**I impulse = m x (vf - vi)**

**Vf = 1.0 m/s**

### 1 Answer | Add Yours

The equation for impulse is correct; however, I have a few comments about the solution:
If this equation is applied to the player-dummy system, then the net force is 0 (dummy and player push each other with equal and opposite forces, according to the thirde Newton's Law.) The change of momentum should be 0 and the momentum conserved. This means that if the runner loses some of the momentum, the dummy will acquire some, but we cannot calculate how much because we do not know the mass of the dummy or its acquired velocity.
If this equation is applied to just the runner, again we cannot calculate the impulse, and thus change of momentum, because we do not know the force with which the dummy pushes the runner.
Where is I = 360 N*s came from ???
I believe more information is needed to solve this problem.

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