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 calculate the initial momentum of the putty and the cart and then compare it to the work done by the friction force.
Let's start by calculating the initial momentum of the putty. Momentum (p) is defined as the product of an object's mass (m) and its velocity (v). The mass of the putty is given as 2.3 grams, which is equal to 0.0023 kg, and the velocity is given as 4.2 m/s.
Momentum of putty = mass of putty * velocity of putty
p_putty = 0.0023 kg * 4.2 m/s
p_putty ≈ 0.00966 kg⋅m/s
Next, we need to calculate the initial momentum of the cart. The mass of the cart is given as 225 grams, which is equal to 0.225 kg. Since the cart is initially at rest, its initial velocity is 0 m/s.
Momentum of cart = mass of cart * velocity of cart
p_cart = 0.225 kg * 0 m/s
p_cart = 0 kg⋅m/s
Now, let's calculate the work done by the friction force on the cart. The work done (W) is given by the equation: W = force * distance * cos(θ), where force is the friction force between the cart and the track, distance is the length of the track, and cos(θ) is the cosine of the angle of the track.
The friction force can be calculated using the normal force (N) between the cart and the track. The normal force is equal to the weight of the cart, which is given by the equation: N = mass of cart * gravitational acceleration (g). The gravitational acceleration is approximately 9.8 m/s^2.
Normal force (N) = mass of cart * gravitational acceleration
N = 0.225 kg * 9.8 m/s^2
N ≈ 2.205 N
Friction force (F) = coefficient of friction (μ) * normal force
Since the coefficient of friction is not given, let's assume it to be 0.2 (a reasonable assumption for this scenario).
F = 0.2 * 2.205 N
F ≈ 0.441 N
Now, we can calculate the work done by the friction force.
W = force * distance * cos(θ)
W = 0.441 N * 1.5 m * cos(25°)
W ≈ 1.86 J
Since work is equal to the change in kinetic energy, we can use the work-energy principle to find the change in velocity of the cart. The initial kinetic energy of the system is equal to the sum of the kinetic energy of the putty and the cart, which is given by the equation: KE = (1/2) * mass * velocity^2.
Initial kinetic energy of system = KE_putty + KE_cart
KE_system = (1/2) * (mass_putty + mass_cart) * velocity^2
KE_system = (1/2) * (0.0023 kg + 0.225 kg) * (4.2 m/s)^2
KE_system ≈ 0.099 J
The change in kinetic energy is given by: ΔKE = KE_final - KE_initial
ΔKE = -W (since work done by friction is negative)
ΔKE = -1.86 J
Finally, we need to calculate the final velocity of the cart. The final kinetic energy of the system is zero because both the putty and the cart come to rest.
Final kinetic energy of system = 0
Using the equation for kinetic energy, we can solve for the final velocity of the cart:
(1/2) * (0.0023 kg + 0.225 kg) * (final velocity)^2 = 0
Simplifying:
(0.2273 kg) * (final velocity)^2 = 0
Since the mass and final velocity are both non-zero, the equation indicates that the final velocity of the cart is zero. Therefore, the cart will not reach the end of the track before it stops and slides back down.
In summary, the calculations show that the initial momentum of the putty and the cart is not enough to overcome the work done by the friction force, resulting in the cart coming to a stop before reaching the end of the track.