Impulse graph

[Halcyon32]

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View attachment 195925

Can you guys explain this for me? Answer is D and the only explanation is the formula Ft=m delta v but how do we figure out what delta v is?

 

[Halcyon32]

i don't know why the graph didn't show up but here it is

 

[sazerac]

Delta V is the answer "how fast".

You know F(average) is 2.5. M is 0.2. T is 20. Solve

 

[popopopop]

My approach on this is that F = ma = m(deltaV)/t, so F(t) = m(deltaV). That's how they came up with that formula.

From Force of 0->5 and time of 10, solving for deltaV = 250 m/s. Because the graph is congruent, going from F of 5->0 at time of 20 meant delvaV's magnitude must be 250 m/s also.

 

[Halcyon32]

Delta V is the answer "how fast".

You know F(average) is 2.5. M is 0.2. T is 20. Solve

Sorry, I meant to say how do we figure out what F is? How did you get 2.5?

 

[popopopop]

Impulse = change in momentum = J = F(deltaT) = deltaP

Finding J(impulse) means finding area under the triangle from 0-20 seconds so F = J/(deltaT).

Half the triangle = 1/2 (10 x 5) = 25. 25 x 2 = 50, the area of the triangle from 0-20 seconds. 50/20 seconds = 2.5 N

 

[Halcyon32]

Thank you! You're the best