Why Is My Calculation of Impulse Twice the Correct Value?

  • Thread starter Thread starter SmokeyBear
  • Start date Start date
  • Tags Tags
    Collision Impulse
AI Thread Summary
The discussion centers on calculating the impulse experienced by a male bighorn sheep during a collision. The initial calculation yielded an impulse of 826.2 Ns, while the correct value is 413 Ns, as determined by finding the area under the acceleration-time graph. The discrepancy arises because the acceleration is not constant, necessitating the use of average acceleration for accurate calculations. The correct approach involves using the average acceleration over the collision duration. Understanding this concept resolves the confusion regarding the doubled impulse value.
SmokeyBear
Messages
2
Reaction score
0
When a male bighorn sheep runs head-first into another male, the rate at which its speed drops to zero is dramatic. Figure 9-11 gives a typical graph of the acceleration a versus time t for such a collision, with the acceleration taken as negative to correspond to an initially positive velocity. The peak acceleration has magnitude 34 m/s2 and the duration of the collision is 0.27 s. Assume that the sheep's mass is 90.0 kg. What are the magnitudes of the impulse and average force due to the collision?

33o2xdk.jpg



I know impulse is equal to the change in momentum (J = mΔv). So first I found vo by using v = vo + at. After plugging in the numbers, I got vo = -9.18 m/s. I then plugged this into the formula J = mΔv = m(v - vo), where v = 0 m/s, to get 826.2 Ns as the impulse. Unfortunately, the book has exactly half of that as the answer (413 Ns) as the answer. The way they did it was finding the area between the plot and the axis of Figure 9-11, then multiplying it by 90 kg. I understand why this works, but what I don't understand is how my answer comes out to be twice as large.

Any help is greatly appreciated. Thanks.
 
Physics news on Phys.org
SmokeyBear said:
I understand why this works, but what I don't understand is how my answer comes out to be twice as large.
The acceleration is not constant. Use the average acceleration in your formula for speed.
 
How did I miss that? Thanks a lot!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
13
Views
8K
Replies
4
Views
2K
Replies
12
Views
2K
Replies
1
Views
2K
Replies
3
Views
5K
Replies
6
Views
7K
Back
Top