OK, let us define mass and weight...
Mass = mass, m
Weight = mass multiplied by gravity, mg
When we weigh something on the earth we weigh its weight. For example - take an apple in your hand - what you are feeling is its
weight - mg, similarly if you use a spring balance (or any balance) to weight it the extension of the spring is determined by its
weight- mg (Hooke's law). However we conveniently refer to it weight in kg, since the force of gravity is (fairly) uniform across the whole Earth.
This is where people get confused. What we are really saying is that the object has a mass of 605 kg (in terms of the F1 car) and weights ~ 5935 N.
Newton's laws apply to mass, not weight.
Jumping (simplified) -
A 50 kg man jumps up with a force of 2500N applied for 0.1 second (as soon as he leaves the ground he can no longer exert a force on it).
As already covered F= ma
Impulse is also governed by Newton's first law -
I = Fdt
I = 250
Impulse also equals the change in momentum -
I = delta p = m delta v
Since we know the mass (50kg) we can determine the change in velocity.
delta v = I/m
delta v = 5 m/s
So the man leaves the ground at 5 m/s
Of course as soon as the man leaves the ground gravity begins to slow him down.
To find how high he jumps we need to determine a few things -
At the maximum height he will be instantaneously at rest - velocity = 0
v = u +at
0 = 5 + at
a = gravity
a = -9.81ms^-2 (Earth, deceleration)
-5 = -9.81t
t= ~0.5 s
thus the man jumps -
s = ut + 0.5a(t^2)
s = 5x0.5 + 0.5*-9.81*0.25
s = 2.5 - 1.23
s = 1.27 m
So on the Earth the man manages to reach 1.27 m off the ground.
What is the story on the Moon??
l = lunar gravity = 1.622 ms^-2
a = l
insert into above equations -
v = u +at
0 = 5 + (-1.622)t
t = 5/1.622
t = ~3 sec
s = ut + 0.5a(t^2)
s = 5*3 + 0.5*-1.622*9
s = 15 - 7.3
s = 7.7 m
Summary -
A man that can jump 1.27 m on Earth, can jump 7.7 m on the moon, while exerting the same force, due to the lesser gravity.
So a poor high jumper on earth is an excellent high jumper on the moon
The situation will be the same for the golf ball, as for a given vertical force the ball will have a longer flight time, enabling it to travel further horizontally.
For a race car on earth - again simplified to assume no slippage and sufficient traction to put the power down.
F = 12100 N
Mass = 605 kg
g = 9.81 ms^-2
F = ma
12100 = 605a
a = 20 ms^-2
The race car accelerates at 20 ms^-2 (or approximately 2 g) from a standstill.
On the moon -
F = 12100 N
l = 1.622 ms^-2
F = ma
12100 = 605a
a = 20 ms^-2
On the moon the car accelerates at 20ms^-2 (or ~12 l), again from a standstill.
Edit: Rolling resistance is complicated
Originally posted by: Pacfanweb
On Earth, gravity helps produce rolling resistance. Tires have more resistance because they are being pushed to the ground with more weight. So you have to overcome all that, plus the mass of the car.
I don't see how the car on the Moon couldn't be at least SOMEWHAT quicker, in the circumstances I outlined.
Yes rolling resistance is determined by (amongst other things) weight.
Rolling resistance is determined by the rolling resistance coefficient (Crr) and the normal force applied (force applied at 90 degrees to the surface).
F = Crr x Nf
So assuming the Crr is the same the difference in rolling resistance between a car on Earth and the Moon is 181 N. Which is sod all.
But I have already said that...
Originally posted by: PlasmaBomb
the lesser force should result in a slightly higher top speed due to less rolling resistance
The problem is with less weight and the same aero drag the speed at which you can deploy full power will be higher (due to less grip). This will kill you on the start and also in hairpins.