A student flings a 2.3g ball of putty at a 225g cart sitting on a slanted air track that is 1.5m long. The track is slanted at an angle of 25 degrees with the horizontal. If the putty is travelling at 4.2 m/s when it hits the cart, will the cart reach the end of the track before it stops and slides back down? Support your answer with calculations.

To determine whether the cart will reach the end of the track before it stops and slides back down, we need to compare the initial kinetic energy of the ball of putty with the potential energy the cart gains as it moves up the track, as well as the energy lost due to friction.

Let's calculate the initial kinetic energy of the ball of putty. The formula for kinetic energy is:

K = 0.5 * m * v^2

Where:
K is the kinetic energy
m is the mass of the object
v is the velocity of the object

Given that the mass of the putty ball is 2.3g (0.0023kg) and its velocity is 4.2 m/s, we can now calculate its kinetic energy.

K = 0.5 * 0.0023kg * (4.2 m/s)^2

K ≈ 0.02037 J (rounded to 5 decimal places)

Now let's determine the potential energy gained by the cart as it moves up the track. The formula for potential energy is:

PE = m * g * h

Where:
PE is the potential energy
m is the mass of the object
g is the acceleration due to gravity (approximately 9.8 m/s^2)
h is the height or vertical distance traveled

To find the height, we need to break down the force components acting on the cart. The gravitational force acting on the cart in the downward direction can be split into two components: one parallel to the track and one perpendicular to it. The parallel component does not contribute to the cart's potential energy as it acts in the same direction as the cart's motion. The perpendicular component is responsible for the cart moving up the inclined track.

The perpendicular component of the gravitational force can be calculated using the equation:

F_perpendicular = m * g * sin(theta),

Where:
F_perpendicular is the force perpendicular to the incline
m is the mass of the cart
g is the acceleration due to gravity
theta is the angle of inclination

Given that the mass of the cart is 225g (0.225kg) and theta is 25 degrees, we can calculate the force perpendicular to the incline:

F_perpendicular = 0.225kg * 9.8 m/s^2 * sin(25 degrees)

F_perpendicular ≈ 0.10091 N (rounded to 5 decimal places)

Now, we can find the height h by dividing the work done by the force perpendicular to the incline:

W = F_perpendicular * h

We know that work done is equal to the potential energy, so we have:

PE = F_perpendicular * h

Rearranging the equation, we get:

h = PE / F_perpendicular

Substituting the values, we find:

h = 0.02037 J / 0.10091 N ≈ 0.20215 m (rounded to 5 decimal places)

Now that we have the height, let's determine the time it takes for the cart to reach the end of the track. We can use the following formula for the time of flight:

t = sqrt((2 * h) / g)

Where:
t is the time of flight
h is the height or vertical distance traveled
g is the acceleration due to gravity (approximately 9.8 m/s^2)

Substituting the values, we find:

t = sqrt((2 * 0.20215 m) / 9.8 m/s^2)

t ≈ 0.20325 s (rounded to 5 decimal places)

Finally, let's calculate the distance traveled by the cart during this time. We can use the formula:

d = v * t

Where:
d is the distance traveled
v is the velocity of the cart
t is the time of flight

Substituting the values, we find:

d = 4.2 m/s * 0.20325 s

d ≈ 0.85485 m (rounded to 5 decimal places)

Comparing the calculated distance of approximately 0.85485 m to the length of the track, which is 1.5 m, we can see that the cart will not reach the end of the track before it stops and slides back down.