In procedure 1: suppose the hanging mass is
mH = 203 g and the mass of the cart is
mC = 1286 g.
a) Assume there is no friction anywhere in the system, and the system starts at rest.
i. Find the magnitude of the acceleration of the masses.
a = m/s2
ii. Find the magnitude of the tension in the string.
T = N
iii. If the track has length L = 2.79 m, find vf, the speed of the cart just before it hits the barrier.
v = m/s
iv. Find tf, the time it takes the cart to travel the 2.79 m to hit the barrier.
tf = s
To find the answers to the given questions, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
a) To find the magnitude of the acceleration of the masses (i), we can use the equation:
net force = (mH + mC) * a
Since there is no friction in the system, the only force causing the acceleration is the force due to gravity acting on the hanging mass. This force is given by:
force of gravity = mH * g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
So the net force equation becomes:
mH * g = (mH + mC) * a
Simplifying the equation and solving for a, we get:
a = (mH * g) / (mH + mC)
Now we can substitute the given values:
mH = 203 g = 0.203 kg (converting grams to kilograms)
mC = 1286 g = 1.286 kg
g = 9.8 m/s^2
Plugging these values into the equation, we can calculate the acceleration:
a = (0.203 kg * 9.8 m/s^2) / (0.203 kg + 1.286 kg)
a ≈ 1.527 m/s^2
Therefore, the magnitude of the acceleration of the masses is approximately 1.527 m/s^2.
ii) To find the magnitude of the tension in the string, we can use the equation:
Tension = mH * g - mH * a
Since we have already calculated the values for mH, g, and a, we can substitute them into the equation:
Tension = (0.203 kg * 9.8 m/s^2) - (0.203 kg * 1.527 m/s^2)
Tension ≈ 1.989 N
Therefore, the magnitude of the tension in the string is approximately 1.989 N.
iii) To find the speed of the cart just before it hits the barrier, we can use the equation:
vf^2 = vi^2 + 2 * a * d
In this case, the initial velocity (vi) is zero, as the system starts at rest. The final velocity (vf) is what we need to find, and the distance traveled (d) is given as L = 2.79 m. We already know the value of acceleration (a) from part (i).
Plugging the given values into the equation, we can solve for vf:
vf^2 = 0 + 2 * (1.527 m/s^2) * (2.79 m)
vf ≈ √(2 * (1.527 m/s^2) * (2.79 m))
Calculating this expression, we find:
vf ≈ 4.34 m/s
Therefore, the speed of the cart just before it hits the barrier is approximately 4.34 m/s.
iv) To find the time it takes the cart to travel the 2.79 m to hit the barrier, we can use the equation:
d = vi * t + (1/2) * a * t^2
As the system starts at rest, the initial velocity (vi) is zero. We are given the distance traveled (d) as L = 2.79 m and the value of acceleration (a) from part (i). We need to find the time (t).
Plugging the given values into the equation, we can solve for t:
2.79 m = 0 + (1/2) * (1.527 m/s^2) * t^2
2.79 m = 0.7635 m/s^2 * t^2
Rearranging the equation, we get:
t^2 = (2.79 m) / (0.7635 m/s^2)
t ≈ √((2.79 m) / (0.7635 m/s^2))
Calculating this expression, we find:
t ≈ 2.22 s
Therefore, the time it takes the cart to travel the 2.79 m to hit the barrier is approximately 2.22 seconds.