Problem 28 - University Physics with Modern Physics

Question

You are standing on a large sheet of frictionless ice and holding a large rock. In order to get off the ice, you throw the rock so it has velocity 12.0 \(\mathrm{m} / \mathrm{s}\) relative to the earth at an angle of \(35.0^{\circ}\) above the horizontal. If your mass is 70.0 \(\mathrm{kg}\) and the rock's mass is \(15.0 \mathrm{kg},\) what is your speed after you throw the rock? (See Discussion Question \(\mathrm{Q} 8.7 . )\)

Step-by-Step Solution

Answer

Your speed after throwing the rock is approximately 2.09 m/s.

Step 1: Identify Known Variables

List out the known values: velocity of the rock relative to the earth, \(v_r = 12.0\, \text{m/s}\), angle \(\theta = 35.0^\circ\), your mass \(m_y = 70.0\, \text{kg}\), and the rock's mass \(m_r = 15.0\, \text{kg}\).

Step 2: Apply Conservation of Momentum

Since the system is isolated and there are no external forces, the linear momentum before and after throwing the rock must be conserved. Initially, both you and the rock are at rest, so the total momentum is zero. This implies: \[ m_y \cdot v_y + m_r \cdot v_{r,x} = 0 \] where \(v_{r,x}\) is the horizontal component of the rock's velocity.

Step 3: Calculate Horizontal Component of Rock's Velocity

Find the horizontal component of the rock's velocity using the formula: \[ v_{r,x} = v_r \cdot \cos(\theta) \]Substitute the known values: \[ v_{r,x} = 12.0 \cdot \cos(35.0^\circ) \] Calculate this to get the horizontal velocity of the rock.

Step 4: Solve for Your Speed

Rearrange the momentum conservation equation to find your speed, \(v_y\):\[ v_y = \frac{-m_r \cdot v_{r,x}}{m_y} \] Calculate \(v_y\) using the value of \(v_{r,x}\) obtained in the previous step.

Key Concepts

Frictionless Surface

Imagine standing on a large sheet of ice with absolutely no friction. On a frictionless surface, there's nothing to slow you down or stop your movement. In physics, this means that any force you apply to something will have an equal and opposite reaction.

For this exercise, when you throw a rock while standing on this frictionless ice, the push against the rock pushes you in the opposite direction. Because there's no friction, this movement is not opposed by any forces. This situation is an ideal thought experiment often used to help understand fundamental concepts like Newton's third law of motion.

In essence, the idea of a frictionless surface helps simplify problems by eliminating the complexity of frictional forces. It allows the principle of conservation of momentum to be clearly observed.

Momentum

Momentum is a fundamental concept in physics, depicted as the product of an object's mass and its velocity. Conservation of momentum tells us that in a closed system—that is, one with no external forces—the total momentum remains constant.

In the exercise above, before you throw the rock, both you and the rock are stationary on the ice. Thus, your combined momentum is zero. When you throw the rock, you both gain momentum in opposite directions.

Mathematically, this is expressed as: - Your momentum (\(m_y \cdot v_y\)) - Rock's momentum (\(m_r \cdot v_{r,x}\))

Since they must sum to zero, the momentum gained by the rock is equal and opposite to the momentum you gain while moving. This is pivotal to solving and understanding the problem at hand.

Projectile Motion

When you throw an object like the rock in this scenario, it becomes a projectile. Projectile motion describes the curved path that an object follows when it's thrown or propelled through the air.

The initial velocity with which you throw the rock, split into horizontal and vertical components, determines its trajectory.

While on a frictionless ice surface, only the horizontal component is relevant for finding your speed because the surface doesn't elevate your movement. This is because the vertical component doesn't affect your momentum on the ice, contributing only to the rock's airborne path. Understanding projectile motion helps decipher how objects move and interact upon thrusting through their own trajectories.

Velocity Components

Whenever an object is thrown at an angle, its velocity can be split into two parts: horizontal and vertical. The horizontal component affects the motion parallel to the ground, while the vertical component influences the motion perpendicular to the ground.

To find the horizontal component of the rock’s velocity we use the cosine of the angle of projection: - \(v_{r,x} = v_r \cdot \cos(\theta)\)

By splitting into components, you can analyze each part separately. Although the vertical velocity changes as the rock moves through the air, the horizontal component remains constant in a frictionless environment. This helps in solving for your speed on the ice because only the horizontal momentum influences your movement, providing a clear path to the solution of what speed you move at after releasing the rock.