View Full Version : Calculating the final speed using Force/Time Graph
1. The problem statement, all variables and given/known data
A golfer strikes a stationary golf ball, and the force applied by the club on the ball varies as shown in the graph below. Use this graph to determine the final speed of the golf ball. The mass of the golf ball is 0.1kg
2. Relevant equations
Area of triangle = 1/2bh (then times by two for the full area.) Or AvgFt=m(v-u)
3. The attempt at a solution
I tried avgF*t= m(v-u) and put 100*0.04=0.1(v-0) = 4/0.1 = 40m/s
also tried Area of Triangle = 1/2bh = 0.5*0.04*100 = 2 Then 2*2 = 4m/s to get the full triangle.
Windowmaker
Oct6-11, 06:26 PM
Final speed? As in t= 80?
Welcome to PH Forums.
The average Force during the 80 milliseconds of contact is 50 Newtons.
What does the Impulse - Momentum Theorem say? -- or -- How is impulse related to momentum?
it says Time of Contact / ms so I assume it's milliseconds.
Welcome to PH Forums.
The average Force during the 80 milliseconds of contact is 50 Newtons.
What does the Impulse - Momentum Theorem say? -- or -- How is impulse related to momentum?
Impulse = Momentum?
Impulse = Momentum?
Not quite right. Look again.
Not quite right. Look again.
Impulse = Change in Momentum.
Impulse = Change in Momentum.
Yes, that's it. Although, in this case, the ball starts out having zero momentum.
Yes, that's it. Although, in this case, the ball starts out having zero momentum.
Ok so how did you get this? Im confused? Sorry about this.. what calculation would I do?
The impulse is the area under the graph, that's the area of the triangle. You have that right as 4 N-s = 4 kg-m/s .
You answer of 40 m/s also looks to be correct.
The impulse is the area under the graph, that's the area of the triangle. You have that right as 4 N-s = 4 kg-m/s .
You answer of 40 m/s also looks to be correct.
So which is the actual answer? you wrote 50 above which kinda confused me.
What did I say about 50 Newtons ? I said it was the average force during the 80 ms of contact. That's all
What did I say about 50 Newtons ? I said it was the average force during the 80 ms of contact. That's all
Ok so the formula I'd use and the solution is what?
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