A 3kg. package slide down a smooth 30 degrees incline and land on a 10 kg. cart that is at rest.Determine the velocity of the cart and the package immediately after the box land on the cart. The package is released from rest 5m up the incline.
h = 5*sin30 = 2.5 m.
V^2 = Vo + 2g*h.
V^2 = 0 + 19.6*2.5 = 49.
V = 7 m/s.
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To determine the velocity of the cart and the package immediately after the box lands on the cart, we can use the principle of conservation of energy.
Step 1: Calculate the potential energy of the package before it is released.
Potential energy (PE) = mass (m) × gravitational acceleration (g) × height (h)
PE = 3 kg × 9.8 m/s^2 × 5 m
PE = 147 J
Step 2: Calculate the potential energy of the package after it lands on the cart.
Since the cart is at rest, the final potential energy of the system (package + cart) is zero.
Step 3: Calculate the kinetic energy of the package when it lands on the cart.
Since the potential energy is converted into kinetic energy, we can equate the two:
PE = KE
147 J = KE
Step 4: Calculate the kinetic energy of the package.
KE = (1/2) × mass (m) × velocity (v)^2
Step 5: Substitute the mass of the package into the equation.
147 J = (1/2) × 3 kg × v^2
Step 6: Solve for the velocity of the package.
v^2 = (2 × 147 J) / 3 kg
v^2 = 98 J / 3 kg
v^2 ≈ 32.67
v ≈ √32.67
v ≈ 5.71 m/s
So, the velocity of the cart and the package immediately after the box lands on the cart is approximately 5.71 m/s.
To determine the velocity of the cart and the package immediately after the box lands on the cart, we need to apply the principles of conservation of energy and momentum.
1. Firstly, we need to calculate the potential energy of the package when it is released from rest 5m up the incline.
- The formula for potential energy is given by PE = mgh, where m is the mass (3 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (5 m).
- Thus, PE = (3 kg)(9.8 m/s^2)(5 m) = 147 J
2. Next, we need to determine the velocity of the package just before it lands on the cart.
- Since the incline is smooth and there is no friction, the package's potential energy at the top of the incline will be converted entirely into kinetic energy just before it lands.
- Therefore, KE = PE = 147 J
- The formula for kinetic energy is given by KE = 0.5mv^2, where m is the mass (3 kg) and v is the velocity of the package.
- Substituting the values, 147 J = 0.5(3 kg)v^2
- Solving for v, we find v^2 = (2 * 147 J) / (3 kg) = 98 m^2/s^2
- Taking the square root of both sides, v ≈ 9.899 m/s (rounded to three decimal places)
3. Now, we can move on to calculating the velocity of the cart and the package immediately after the box lands on the cart.
- By applying the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
- The momentum of an object is given by p = mv, where m is the mass and v is the velocity.
- Before the collision, the momentum of the package is (3 kg)(9.899 m/s) = 29.697 kg·m/s (rounded to three decimal places).
- Since the cart is at rest before the collision, its momentum is zero.
- After the collision, the combined momentum of the cart and package is conserved and equal to 29.697 kg·m/s.
- Let v' be the velocity of the cart and package after the collision.
- Thus, (10 kg)(v') = 29.697 kg·m/s
- Solving for v', we find v' ≈ 2.970 m/s (rounded to three decimal places)