- #1
1831
- 13
- 0
Two air-track carts, one on the left with mass 100g and one on the left with mass 300g) are sliding to the right at 1.0 m/s. There is a spring between them that has a spring constant of 100N/m and is compressed 4.2 cm. There is a string that holds the two carts together. The carts slide past a flame that burns through the string holding them together.
(After the flame burns the string)...
Q1: What is the speed of 100g cart?
Q2: What is the direction of the motion of 100g cart? I got to the left
Q3:What is the speed of 300g cart?
Q4:What is the direction of the motion of 300g cart? I got to the right
Here is how I started off for Q1 and Q3:
F=-k[tex]\Delta[/tex]s
F=(100)(.042) = 4.2 N
@Q1 I drew my force diagram, so the force by the spring on the 100g block points in the negative direction, so I got:
[tex]\sum[/tex]Fx = -Fspring = ma
-4.2N = (.1kg)(acceleration)
acceleration = -42m/s^2
Then, I used kinematics to solve for final velocity (vf)
Vf^2 = Vi^2 + 2a[tex]\Delta[/tex]s
since the spring is compressed 4.2cm, then the distance for it to return to equilibrium is .021m on each side...so
Vf^2 = 1^2 +2(-42)(.021)
vf^2=-.764
Now I'm stuck, because I don't want to take the square root of a negative...which makes me think I've done something wrong.
I will follow the same procedure for Q3...once I figure out what I'm doing wrong in Q1.
Please help me where I've messed up.
Thanks!
(After the flame burns the string)...
Q1: What is the speed of 100g cart?
Q2: What is the direction of the motion of 100g cart? I got to the left
Q3:What is the speed of 300g cart?
Q4:What is the direction of the motion of 300g cart? I got to the right
Here is how I started off for Q1 and Q3:
F=-k[tex]\Delta[/tex]s
F=(100)(.042) = 4.2 N
@Q1 I drew my force diagram, so the force by the spring on the 100g block points in the negative direction, so I got:
[tex]\sum[/tex]Fx = -Fspring = ma
-4.2N = (.1kg)(acceleration)
acceleration = -42m/s^2
Then, I used kinematics to solve for final velocity (vf)
Vf^2 = Vi^2 + 2a[tex]\Delta[/tex]s
since the spring is compressed 4.2cm, then the distance for it to return to equilibrium is .021m on each side...so
Vf^2 = 1^2 +2(-42)(.021)
vf^2=-.764
Now I'm stuck, because I don't want to take the square root of a negative...which makes me think I've done something wrong.
I will follow the same procedure for Q3...once I figure out what I'm doing wrong in Q1.
Please help me where I've messed up.
Thanks!