06 Feb 10
Originally posted by uzless'impact' is a bit of a wishy-washy word but lets assume you mean the energy of the impact.
You drop a 300 pound ball to the ground from a height of 1cm.
You then drop a 100 pound ball to the ground from a height of 10cm.
If you had a device to measure the impact on the ground, which would have a higher impact, the 100 pound ball or the 300 pound ball? Show your calcs.
since PE is proportional to weight and altitude the 100 pound ball will have more energy on impact.
Precisely the lighter ball has 10 metre-pounds of energy
The heavier ball has 3 metre-pounds of energy
06 Feb 10
Originally posted by uzlessDrawing a free body diagram around a ball, we see that the only force acting on the ball is the force of gravity acting in the downward direction:
You drop a 300 pound ball to the ground from a height of 1cm.
You then drop a 100 pound ball to the ground from a height of 10cm.
If you had a device to measure the impact on the ground, which would have a higher impact, the 100 pound ball or the 300 pound ball? Show your calcs.
F = ma = -mg
Dividing through by the mass "m", we see that:
a = -g
And therefore:
y" = -g
Integrating both sides, and assuming the ball is initially at rest, we get:
y' = -gt + 0 = -gt
Integrating both sides again, and assuming a height of "h", we get:
y = -(1/2)gt^2 + h
The ball will increase in velocity as it is dropped until it reaches a maximum just before impact. Letting y = 0 and solving for t, we have:
(1/2)gt^2 = h
t = SQRT(2h/g)
Subbing this value into the expression for the velocity, we have:
y' = -g*SQRT(2h/g) = -SQRT(2gh)
The maximum momentum of the ball is the product of this and its mass:
p = mv = -m*SQRT(2gh)
Subbing in the values provided in the question, we find the following:
m1 = 300 lbs = 300/2.2 kg =~ 136.36 kg
h1 = 1 cm = 0.01 m
g = 9.81 m/s^2
p1 = -(136.36)*SQRT(2*9.81*0.01) = 60.4 kg*m/s
m2 = 100 lbs = 100/2.2 kg =~ 45.45 kg
h2 = 10 cm = 0.1 m
g = 9.81 m/s^2
p2 = -(45.45)*SQRT(2*9.81*0.1) = 63.7 kg*m/s
Therefore the momentum of the second ball will be higher at impact. Now, the force of impact depends on a number of factors including the material properties of the ball, the material properties of the impact surface, the momentum of the ball, the angle of impact, etc... but we can make the simplifying assumption that both balls have the same material properties (including density). This assumption simplifies things because if they both have the same density, the second ball will be smaller. This means that the force of the second ball's impact will be spread over less area, resulting in greater pressure at impact. If the definition of "greater impact" is greater penetrating power, then the second ball clearly has greater impact.
07 Feb 10
Originally posted by PBE6So suppose you drop the 300 kg one from 10 Cm, would it have ten times the impact effect? Same size, same density. Or would it have the square of the impact energy of the first drop?
Drawing a free body diagram around a ball, we see that the only force acting on the ball is the force of gravity acting in the downward direction:
F = ma = -mg
Dividing through by the mass "m", we see that:
a = -g
And therefore:
y" = -g
Integrating both sides, and assuming the ball is initially at rest, we get:
y' = -gt + 0 = -gt
Integra ...[text shortened]... impact" is greater penetrating power, then the second ball clearly has greater impact.
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07 Feb 10
1 edit
Originally posted by sonhouseIn that particular case it would increase it momentum by a factor of sqrt(10).
So suppose you drop the 300 kg one from 10 Cm, would it have ten times the impact effect? Same size, same density. Or would it have the square of the impact energy of the first drop?
heres how
P(h) = (2g/h)^(1/2)........eq1
That is the general equation for this objects momentum give the initial height
its original height was 1 cm, and
10*(1 cm) = 10 cm = 10*h
plug it in
P(10*h) = (2g/(10*h))^(1/2)
............= [(2g/(h))^(1/2)]*(1/(10^(1/2)))
so multiplying both sides of the equation by (10^(1/2)) we obtain
(10^(1/2))(P(10h)) = (2g/h)^(1/2)..........eq 2
now you can compare eq1 to eq 2 because both righthand sides of the equation are equal( ie they both equal (2g/h)^(1/2))
so
P(h) = (sqrt(10))P(10h)
what this says is that by multiplying the height by any factor "x" you will increase the momentum by sqrt(x)
note: the relationship is only this simple if you " multiply h by a factor x", if you add a term to h the relationship is not as simple(ie how much will the momentum increase by if you add so many cm to the original height).
07 Feb 10
Originally posted by joe shmoSo you have to square the distance to get 10 times the energy of impact, so 1 Cm to 10 gives you 3 and change times at 10, and only 10 times the energy of impact at 100 cm. 100 times the energy would mean you have to drop it from 10,000 cm, right?
In that particular case it would increase it momentum by a factor of sqrt(10).
heres how
P(h) = (2g/h)^(1/2)........eq1
That is the general equation for this objects momentum give the initial height
its original height was 1 cm, and
10*(1 cm) = 10 cm = 10*h
plug it in
P(10*h) = (2g/(10*h))^(1/2)
............= [(2g/(h))^(1/2)]*(1/(1 ...[text shortened]... f you " multiply h by a factor x", if you add a term to h the relationship is not as simple.
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07 Feb 10
Originally posted by sonhousenot in general, ( it may turn out to be the case with these particular numbers), but here is the equation you must solve to find the answer you seek.
So you have to square the distance to get 10 times the energy of impact, so 1 Cm to 10 gives you 3 and change times at 10, and only 10 times the energy of impact at 100 cm. 100 times the energy would mean you have to drop it from 10,000 cm, right?
10*P = sqrt(2g/(x*h)) solve this for x, and remember that P = Sqrt(2g/h)
and once you solve this youll see that if the hieght was originally 2 cm that squaring the eight DOES NOT give you 10 times the momentum, thus your conclusion of squaring the original height is incorrect.
07 Feb 10
Originally posted by joe shmog in this case being gravity I assume, 32 f/sec squared or 9.8 m/s squared.
not in general, ( it may turn out to be the case with these particular numbers), but here is the equation you must solve to find the answer you seek.
10*P = sqrt(2g/(x*h)) solve this for x, and remember that P = Sqrt(2g/h)
and once you solve this youll see that if the hieght was originally 2 cm that squaring the eight DOES NOT give you 10 times the momentum, thus your conclusion of squaring the original height is incorrect.
07 Feb 10
Originally posted by PBE6A smaller ball will not have the same air-friction as the larger ball.
Drawing a free body diagram around a ball, we see that the only force acting on the ball is the force of gravity acting in the downward direction:
F = ma = -mg
Dividing through by the mass "m", we see that:
a = -g
And therefore:
y" = -g
Integrating both sides, and assuming the ball is initially at rest, we get:
y' = -gt + 0 = -gt
Integra ...[text shortened]... impact" is greater penetrating power, then the second ball clearly has greater impact.
But it's either that or a difference in density 🙂
07 Feb 10
Originally posted by deriver69The New York ball hits the ground first although the London ball will be going faster.
my particular favourite ball dropping problem is where two identical balls each with mass 1kg are dropped out of a first floor window at 12.00 GMT. One in New York the other in London, which one hits the ground first?
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08 Feb 10
2 edits
Originally posted by joe shmoWhat im trying to say is that the change in momentum is not only dependent on the original height, but also how the height is changed.
not in general, ( it may turn out to be the case with these particular numbers), but here is the equation you must solve to find the answer you seek.
10*P = sqrt(2g/(x*h)) solve this for x, and remember that P = Sqrt(2g/h)
and once you solve this youll see that if the hieght was originally 2 cm that squaring the eight DOES NOT give you 10 times the momentum, thus your conclusion of squaring the original height is incorrect.
P=sqrt(2g/(h+c)), for example if you add a certain height "c", the new factor by which the momentum increases(using some algebra) is
sqrt(1 + c/h)
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09 Feb 10
Originally posted by wolfgang59I like that answer
'impact' is a bit of a wishy-washy word but lets assume you mean the energy of the impact.
since PE is proportional to weight and altitude the 100 pound ball will have more energy on impact.
Precisely the lighter ball has 10 metre-pounds of energy
The heavier ball has 3 metre-pounds of energy
potential energy (approx)= mass.G.height_above_earth_surface
it's approx because gravity varies with distance, but it's good enough over this distance
So, as 100*10 > 300 * 1, the lighter ball has over 3 times the energy at the point of impact.
To do it another way:
G = A = 9.8 m/s/s
kinetic_energy = 1/2 * mass * velocity**2
velocity on impact = A * T
distance travelled on impact = 1/2 AT**2
Time taken to impact = sqrt(distance*2/A)
velocity at impact = sqrt(A*distance*2)
kinetic_energy_at_impact = 1/2 * mass * A * distance * 2
= mass.G. distance
Again, the lighter ball has over 3 times the energy at the point of impact.