Problem 15 - College Physics

Question

You (mass \(55 \mathrm{~kg}\) ) are riding your frictionless skateboard (mass \(5.0 \mathrm{~kg}\) ) in a straight line at a speed of \(4.5 \mathrm{~m} / \mathrm{s}\) when a friend standing on a balcony above you drops a \(2.5 \mathrm{~kg}\) sack of flour straight down into your arms. (a) What is your new speed, while holding the flour sack? (b) since the sack was dropped vertically, how can it affect your horizontal motion? Explain. (c) Suppose you now try to rid yourself of the extra weight by throwing the flour sack straight up. What will be your speed while the sack is in the air? Explain.

Step-by-Step Solution

Answer

(a) New speed is 4.32 m/s. (b) Vertical motion doesn't affect horizontal speed. (c) Speed returns to 4.5 m/s as total mass decreases.

Step 1: Understand the Conservation of Momentum

Since there is no external horizontal force acting on the system (you, the skateboard, and the sack), the conservation of momentum principle applies. This means the total momentum before and after catching the sack remains the same. Mathematically, this is expressed as:\[ m_1v_1 + m_2v_2 = (m_1 + m_2)v_3 \]Where:- \(m_1\) and \(v_1\) are the mass and velocity of the skateboarder and skateboard: \(60\, \text{kg}\) and \(4.5\, \text{m/s}\).- \(m_2\) and \(v_2\) are the mass and velocity of the sack: \(2.5\, \text{kg}\) and \(0\, \text{m/s}\) (since initially it's dropped vertically).- \(v_3\) is the final velocity.

Step 2: Calculate the New Speed After Catching the Sack

Plug the known values into the conservation of momentum equation. Since \(v_2 = 0\), the equation simplifies to:\[ (60 \times 4.5) + (2.5 \times 0) = (60 + 2.5)v_3 \]This simplifies to:\[ 270 = 62.5v_3 \]To find \(v_3\), divide both sides by 62.5:\[ v_3 = \frac{270}{62.5} \approx 4.32\, \text{m/s} \]

Step 3: Impact of Vertical Drop on Horizontal Motion

The vertical drop of the sack doesn't affect the horizontal motion due to independence of the vertical and horizontal components of motion in physics. The velocity change happens solely due to the added mass affecting the momentum.

Step 4: Calculate Speed After Throwing Sack Up

When you throw the sack straight up, a similar conservation of momentum situation occurs but in reverse. Initially, you are moving at \(v_3 = 4.32\, \text{m/s}\) carrying the sack. When the sack leaves your hands, only your body and skateboard are in motion:\[ (62.5 \times 4.32) = (60)v_f + (2.5 \times 0) \]This simplifies to:\[ 270 = 60v_f \]Therefore, solving for \(v_f\) gives:\[ v_f = \frac{270}{60} = 4.5\, \text{m/s} \]

Step 5: Explanation of Speed Recovery

Throwing the sack upwards does not change the horizontal momentum significantly since the upward motion has no horizontal force component. Thus, your velocity when the sack is in the air returns to the original value of \(4.5\, \text{m/s}\), because you have effectively reduced the total moving mass back to the initial condition.

Key Concepts

Physics Problem Solving

In the realm of physics problem-solving, understanding the context and applying appropriate principles is crucial. This involves identifying what is given, what you need to find, and which laws of physics apply to the scenario. For the skateboarder and flour sack problem, the key principle is the conservation of momentum.
Before solving, it's important to gather all known variables and understand the framework. You weigh 55 kg, and the skateboard weighs 5 kg, adding up to a total mass of 60 kg. You begin moving at a speed of 4.5 m/s. A 2.5 kg sack of flour is dropped vertically into your arms. Here, you must calculate the new speed as these components interact.

• First, identify that no external horizontal forces are acting, which means momentum is conserved in the horizontal direction.
• Establish the conservation of momentum equation by noting the initial and final states of momentum in the system.
• Solve for the desired value, typically involving substituting all known quantities and mathematical manipulation.

By following these steps, tackling physics problems becomes an exercise in logical application and calculation, ensuring you arrive at accurate results.

Momentum in Collision

Momentum is a crucial concept in understanding how objects interact, especially during collisions. It is defined as the product of an object's mass and velocity. The law of conservation of momentum states that in an isolated system, the total momentum before and after a collision is the same as long as no external forces are acting on it.
In this situation, when the sack of flour is caught, it effectively "collides" with the skateboarder in a way that conserves the momentum horizontally. The initial momentum of the system is only carried by the skateboarder and board since the sack is dropped with no horizontal velocity.

• Initial momentum: Only the momentum of the skateboarder and skateboard matters, calculated as the product of 60 kg and 4.5 m/s.
• Final momentum: After catching the sack, the total mass becomes 62.5 kg, and the velocity must be recalculated to maintain the same momentum.

This conservation showcases how adding mass, even from a vertical drop, impacts horizontal velocity through the transfer and conservation of momentum.

Independence of Motion Components

The principle of independence of motion components is another key concept needed to fully understand the skateboard scenario. In physics, motion along different axes (horizontal and vertical) are generally independent of each other. This means that changes in motion in one direction don't affect motion in another direction.
When the sack is dropped vertically, it has no effect on your horizontal velocity. The vertical and horizontal components of motion do not interfere with one another.

• The vertical force of gravity affects only the downward motion of the sack, not the horizontal motion of you and the skateboard.
• Horizontally, your speed changes purely due to the increase in combined mass from catching the sack, not the sack's vertical drop.

Understanding these independent components clarifies why the initial horizontal momentum calculation remains unaltered by the vertical motion, and why strategies like throwing the sack straight up return you to your original velocity.