# Calculating final velocity after a puck is hit

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1. Oct 15, 2016

### kinghunter

1. The problem statement, all variables and given/known data
Bob passes a 0.220kg puck with an initial velocity of 12m/s [South] to Dave. Dave one-times the puck with a force of 368N [N30E] if Daves stick is in contact with the puck for 0.250s what is the final velocity?

2. Relevant equations
i honestly don't know where to start with this one, if you can tell me what i first need to solve for i should be able to go from there

3. The attempt at a solution

Last edited: Oct 15, 2016
2. Oct 15, 2016

### billy_joule

Look in your course notes and/or text book for "Impulse".

As the question is, it can't be solved, we don't know the initial direction of the puck.

3. Oct 15, 2016

### kinghunter

sorry the initial directon is south

4. Oct 15, 2016

### SammyS

Staff Emeritus
Hello kinghunter. Welcome to PF !

What quantities are you given? What do you need to find ?

Consider using the Impulse - Momentum Theorem.

You will need to show an attempt before we can give you any more help.

5. Oct 15, 2016

### kinghunter

i believe the initial velocity when dave hits the puck is 12m/s (carried over from the initial pass), he hits it with a force of 368N [N30E], its in contact with his stick for 0.25s (acceleration period) and the mass of the puck is 0.22kg so what i did is F=ma and a=v/t F=m(vf-vi/t)
vf=F(t)+vi / m
=368N(0.25s)+12m/s / 0.22kg
but then i get an unrealistic number = 472.7m/s

6. Oct 15, 2016

### SammyS

Staff Emeritus
For one thing your units are don't work out in several places, partly because you ignore placing parentheses adequately.

Beyond that: Force, velocity and acceleration are all vector quantities. A force in the east - west direction will only cause acceleration in the east - west direction so only changes the component of velocity in the east - west direction. The component of velocity in the north - south direction will remain unchanged at 12 m/s [south] .