Question

A ramp of mass M is at rest on a horizontal surface. A small cart of mass m is placed at the top of the ramp and released. What are the velocities of the ramp and the cart relative to the ground at the instant the cart leaves the ramp? (Assume the positive direction is to the right. Indicate the direction with the signs of your answers. Use the following as necessary: M, m, g, and h.)

Answer 1

When the small cart is released and leaves the ramp, the velocities of the ramp and the cart can be determined using the principles of conservation of momentum and energy. By equating the initial potential energy of the cart to its final kinetic energy, and the initial momentum of the cart+ramp system to its final momentum, the velocities can be found through simultaneous equations. In an example calculation, when the cart has mass 2 kg, the ramp has mass 10 kg, the height of the ramp is 5 m, and the acceleration due to gravity is 9.8 m/s^2, the cart's velocity relative to the ground is +9.36 m/s to the right, and the ramp's velocity relative to the ground is +1/6 m/s to the right.

Step-by-step explanation:

When the small cart is released and leaves the ramp, the velocities of the ramp and the cart can be determined by using the principles of conservation of momentum and conservation of energy.

1. Using conservation of energy, we can equate the initial potential energy of the cart at the top of the ramp to its final kinetic energy when it leaves the ramp.
2. Using conservation of momentum, we can equate the initial momentum of the cart+ramp system to its final momentum when the cart leaves the ramp.
3. By solving these equations simultaneously, we can find the velocities of the ramp and the cart relative to the ground.

Let's assume the cart moves to the right with velocity v and the ramp moves to the right with velocity V.

Using conservation of energy:

mgh = (m/2)v^2 + (M/2)V^2

Using conservation of momentum:

mvo = (M+m)V

By substituting the value of V from the second equation into the first equation and solving for v, we can find the velocity of the cart relative to the ground at the instant it leaves the ramp. The velocity of the ramp relative to the ground can then be found by substituting the value of V into the second equation.

For example, if the cart has mass m = 2 kg, the ramp has mass M = 10 kg, the height of the ramp is h = 5 m, and the acceleration due to gravity is g = 9.8 m/s^2, we can calculate the velocities as follows:

mgh = (m/2)v^2 + (M/2)V^2

(2)(9.8)(5) = (2/2)v^2 + (10/2)V^2

98 = v^2 + 25V^2

mvo = (M+m)V

(2)o = (10+2)V

2 = 12V

From the second equation, V = 1/6 m/s.

Substituting this value into the first equation:

98 = v^2 + 25(1/6)^2

98 = v^2 + 25/36

v^2 = 98 - 25/36

v^2 = 87.69

v = ±9.36 m/s (taking the positive value since the question specifies the positive direction is to the right)

Therefore, the velocity of the cart relative to the ground at the instant it leaves the ramp is +9.36 m/s to the right, and the velocity of the ramp relative to the ground at the same instant is +1/6 m/s to the right.