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Dynamics
Hey all how u doin. i need help with the following 2 questions.
1. A mass m is given an initial velocity of 'u' in the x-direction. It moves in a medium that resists its motion with a force proportional to the velocity, F = -kv. Find v(x) and how far the mass travels before stopping. Find v(t)
2. A ball is thrown horizontally from the top of 150-ft cliff at A with a speed of 50ft/s and lands at point C. Because of a strong horizontal wind, the ball has a constant acceleration in the negative x-direction. Determine the radius of curvature 'p' of the path of the ball at B, where its trajectory makes an angle 45 degrees with the horizontal. i got "normal acc" to be 38. need to check on tht.
p.s. the attached pic is for question 2
thnk u
1. A mass m is given an initial velocity of 'u' in the x-direction. It moves in a medium that resists its motion with a force proportional to the velocity, F = -kv. Find v(x) and how far the mass travels before stopping. Find v(t)
2. A ball is thrown horizontally from the top of 150-ft cliff at A with a speed of 50ft/s and lands at point C. Because of a strong horizontal wind, the ball has a constant acceleration in the negative x-direction. Determine the radius of curvature 'p' of the path of the ball at B, where its trajectory makes an angle 45 degrees with the horizontal. i got "normal acc" to be 38. need to check on tht.
p.s. the attached pic is for question 2
thnk u
Scroll to see replies
Reply 1
for the first one:
by definition, and by using the chain rule
but is just v, so:
use F=ma and replace a with v*dv/dx, separate variables and integrate to get v as a function of x. it will stop when v = 0 so you can do the second bit. to find v as a function of t, just use the definition a = dv/dt, again separating and integrating.
by definition, and by using the chain rule
but is just v, so:
use F=ma and replace a with v*dv/dx, separate variables and integrate to get v as a function of x. it will stop when v = 0 so you can do the second bit. to find v as a function of t, just use the definition a = dv/dt, again separating and integrating.
Reply 3
not that the people in this forum aren't good, but you're more likely to get a quick response in the maths forum!
i'm not sure about that one.
i'm not sure about that one.
Reply 4
Its certainly a question that is more likely to appear on a Mechanics paper than a recent Physics exam - but there were lots of questions like this once upon a time in A level Physics 
Reply 5
The second part of the question uses the results from the first!
You need to get the horizontal velocity as a function of x
and the vertical velocity as a function in y. eg v^2= 2 g (h - y)
You need to get the horizontal velocity as a function of x
and the vertical velocity as a function in y. eg v^2= 2 g (h - y)
Reply 7
teachercol
Its certainly a question that is more likely to appear on a Mechanics paper than a recent Physics exam - but there were lots of questions like this once upon a time in A level Physics
wish they were like that now - mechanics in physics is too un-mathsy for me!
k0rrupter
Hey all how u doin. i need help with the following 2 questions.
2. A ball is thrown horizontally from the top of 150-ft cliff at A with a speed of 50ft/s and lands at point C. Because of a strong horizontal wind, the ball has a constant acceleration in the negative x-direction. Determine the radius of curvature 'p' of the path of the ball at B, where its trajectory makes an angle 45 degrees with the horizontal. i got "normal acc" to be 38. need to check on tht.
thnk u
2. A ball is thrown horizontally from the top of 150-ft cliff at A with a speed of 50ft/s and lands at point C. Because of a strong horizontal wind, the ball has a constant acceleration in the negative x-direction. Determine the radius of curvature 'p' of the path of the ball at B, where its trajectory makes an angle 45 degrees with the horizontal. i got "normal acc" to be 38. need to check on tht.
thnk u
Have you got a value for d - deceleration from the wind ?
x is displacement in the x-direction - +ve to the right.
y is displacement in the y-direction - +ve in the downwards direction.
v - initial horizontal velocity.
g - accln due to gravity - taken as +ve.
d - decceln due to wind - taken as -ve.
t - time of displacement.
x = v.t - (1/2)dt²
y = (1/2)gt² ----- > t² = 2y/g, and t = √(2y/g)
substituting for t² and t in the 1st eqn,
x = v.√(2y/g) - (1/2)d(2y/g)
dx/dy = (1/2)v(2y/g)^(-1/2)(2/g) - d/g
setting dx/dy = 1 (45 degree angle for trajectory)
(v/g)√(g/2y) = 1 + d/g = (d+g)/g
√(g/2y) = (d+g)/v
2y/g = v²/(d+g)²
y = (1/2)(v²g)/(d+g)²
================
Ok, now substitute for v,g,d into the above to get the value for y when the trajectory is 45 degrees.
For the radius of curvature, use
R = [1 + (y')²]^(3/2) / |d²y/dx²|
Use the above expresion for dx/dy to get the 2nd derivative.
y is displacement in the y-direction - +ve in the downwards direction.
v - initial horizontal velocity.
g - accln due to gravity - taken as +ve.
d - decceln due to wind - taken as -ve.
t - time of displacement.
x = v.t - (1/2)dt²
y = (1/2)gt² ----- > t² = 2y/g, and t = √(2y/g)
substituting for t² and t in the 1st eqn,
x = v.√(2y/g) - (1/2)d(2y/g)
dx/dy = (1/2)v(2y/g)^(-1/2)(2/g) - d/g
setting dx/dy = 1 (45 degree angle for trajectory)
(v/g)√(g/2y) = 1 + d/g = (d+g)/g
√(g/2y) = (d+g)/v
2y/g = v²/(d+g)²
y = (1/2)(v²g)/(d+g)²
================
Ok, now substitute for v,g,d into the above to get the value for y when the trajectory is 45 degrees.
For the radius of curvature, use
R = [1 + (y')²]^(3/2) / |d²y/dx²|
Use the above expresion for dx/dy to get the 2nd derivative.
Reply 13
steve10
what is a ?
acceleration I think (you can tell from the units apart from anything else)
k0rrupter
dont u have to get the tangential acceleration and normal acceleration
Nope!
You're asked for the radius of curvature, for which you need the eqn of the curve (more or less)
You are told the slope at which you are interested, slope = 1.
Then dy/dx = 1.
Fom the eqns of motion - for displacement in x- and y-directions, you can get an (implicit) eqn in x and y.
Use this to get the slope - etc.
You don't need to work out those accelerations.
Ah, that's motion in a circle stuff. I get it now.
Is this q from your work on Motion in a circle ?
It just seems to me to be a kinematics question - using suvat eqns and the like.
Since you are asked for the radius of curvature, then you only need the equation for the path of the ball.
I think my post, #11, will answer your q for you.
Is this q from your work on Motion in a circle ?
It just seems to me to be a kinematics question - using suvat eqns and the like.
Since you are asked for the radius of curvature, then you only need the equation for the path of the ball.
I think my post, #11, will answer your q for you.
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