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raphile
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Hey, I have an answer to this problem but really need someone to look through it and see if I've used the right method. I have an exam in a few days! Thanks :)
A crow drops a mollusc when moving with speed 2m/s in a level flight at a height of 28.8m above rocky ground. By taking the gravitational acceleration to be 9.81m/s^2 and ignoring air resistance, specify the velocity of the mollusc in magnitude and direction on its impact with the ground.
If the mass of the shellfish is 20 grams and the duration of the impact is 0.03s, find the horizontal and vertical components of the average force that acts on the shell if the forward and upward components of the velocity of the mollusc after impact are 0.8m/s and 3m/s respectively.
The horizontal speed of the mollusc remains constant at 2m/s during the fall.
To find the vertical speed, first find the duration of the fall:
using s=ut+(1/2)at^2 we have 28.8 = (1/2*9.81*t^2) => t = 2.423 seconds.
Hence the vertical speed at impact is, using v=u+at: (0+(9.81*2.423)) = 23.77m/s.
So the magnitude of the speed at impact is sqrt(2^2+23.77^2) = 23.855m/s.
And the direction is arctan(23.77/2) = 85.2 degrees below the horizontal.
Now onto the second part...
The acceleration horizontally is (change in velocity divide time) = (0.8-2)/0.03 = -40. Hence the force acting forward is (mass times acceleration) = -0.8N.
The acceleration vertically is, using the same method, (3-(-23.77)/0.03 = 892.33. Hence the force acting upward is (892.33*0.02) = 17.85N.
Is this right? Please do check it, as I'm not convinced by the numbers I've got and my tutor is currently unavailable to help. Thanks.
Homework Statement
A crow drops a mollusc when moving with speed 2m/s in a level flight at a height of 28.8m above rocky ground. By taking the gravitational acceleration to be 9.81m/s^2 and ignoring air resistance, specify the velocity of the mollusc in magnitude and direction on its impact with the ground.
If the mass of the shellfish is 20 grams and the duration of the impact is 0.03s, find the horizontal and vertical components of the average force that acts on the shell if the forward and upward components of the velocity of the mollusc after impact are 0.8m/s and 3m/s respectively.
The Attempt at a Solution
The horizontal speed of the mollusc remains constant at 2m/s during the fall.
To find the vertical speed, first find the duration of the fall:
using s=ut+(1/2)at^2 we have 28.8 = (1/2*9.81*t^2) => t = 2.423 seconds.
Hence the vertical speed at impact is, using v=u+at: (0+(9.81*2.423)) = 23.77m/s.
So the magnitude of the speed at impact is sqrt(2^2+23.77^2) = 23.855m/s.
And the direction is arctan(23.77/2) = 85.2 degrees below the horizontal.
Now onto the second part...
The acceleration horizontally is (change in velocity divide time) = (0.8-2)/0.03 = -40. Hence the force acting forward is (mass times acceleration) = -0.8N.
The acceleration vertically is, using the same method, (3-(-23.77)/0.03 = 892.33. Hence the force acting upward is (892.33*0.02) = 17.85N.
Is this right? Please do check it, as I'm not convinced by the numbers I've got and my tutor is currently unavailable to help. Thanks.
