- #1
Buri
- 273
- 0
Ann (mass 50kg) is standing at the left end of a 15m long 500kg cart that has frictionless wheels and rolls on a frictionless track. Initially Ann and the cart are at rest. Suddenly, Ann starts running along the cart at a speed of 5.0m/s relative to the cart.
How far will Ann have have run relative to the ground when she reached the right end of the cart?Solution
When doing this problem I got the right answer only if I added in a negative sign at the end-hardly the right way do to physics. (First year honors).
This is the way I defined my variables:
Ma=mass of Ann
Mc= mass of cart
Vaf = final velocity of Ann
Vai = initial velocity of Ann
Vcf, Vci defined similarly for the cart
and the equation for relativity x = x' + Vxt
where x = Earth's reference frame
x' = carts reference frame
Vx = relative velocity
t = time
I begin with the equation for the conservation of momentum:
MaVaf + McVcf = MaVai + McVci
However I know once Ann reaches the other side her velocity will be zero, however since she's on the cart she will actually have the velocity of the cart in Earth's reference frame. Therefore the equation above simplifies:
Vf(Ma + Mc) = MaVai
Where we call Vaf = Vcf = Vf and Vci = 0
and I find Vf = 0.45m/s
Here is where my problem starts; Vf is positive! I think it should be negative - but I don't see a way to make it "come out" negative.
ADDING the negative sign and realizing it will take Ann 3s to get across, relativity gives me:
x = x' + Vxt
= 15 - (0.45)(3)
= 13.6m
Which is the correct answer! But I had to ADD the negative, something that I think I shouldn't have to do.
Any help?
How far will Ann have have run relative to the ground when she reached the right end of the cart?Solution
When doing this problem I got the right answer only if I added in a negative sign at the end-hardly the right way do to physics. (First year honors).
This is the way I defined my variables:
Ma=mass of Ann
Mc= mass of cart
Vaf = final velocity of Ann
Vai = initial velocity of Ann
Vcf, Vci defined similarly for the cart
and the equation for relativity x = x' + Vxt
where x = Earth's reference frame
x' = carts reference frame
Vx = relative velocity
t = time
I begin with the equation for the conservation of momentum:
MaVaf + McVcf = MaVai + McVci
However I know once Ann reaches the other side her velocity will be zero, however since she's on the cart she will actually have the velocity of the cart in Earth's reference frame. Therefore the equation above simplifies:
Vf(Ma + Mc) = MaVai
Where we call Vaf = Vcf = Vf and Vci = 0
and I find Vf = 0.45m/s
Here is where my problem starts; Vf is positive! I think it should be negative - but I don't see a way to make it "come out" negative.
ADDING the negative sign and realizing it will take Ann 3s to get across, relativity gives me:
x = x' + Vxt
= 15 - (0.45)(3)
= 13.6m
Which is the correct answer! But I had to ADD the negative, something that I think I shouldn't have to do.
Any help?