Problem 54 - College Physics

Question

A \(0.4 \mathrm{~kg}\) stone is thrown horizontally at a speed of \(20 \mathrm{~m} / \mathrm{s}\) from a \(40-\mathrm{m}\) -tall building. (a) Determine the \(x\) and \(y\) components of the stone's momentum the moment after it is thrown. (b) What are the components of its momentum just before it hits the ground? What impulse did gravity impart to the stone?

Step-by-Step Solution

Answer

(a) \( p_x = 8 \text{ kg m/s} \), \( p_y = 0 \text{ kg m/s} \). (b) \( p_x = 8 \text{ kg m/s} \), \( p_y = 11.2 \text{ kg m/s} \). Impulse: \( 11.2 \text{ kg m/s} \).

Step 1: Determine the Initial Horizontal Momentum

The momentum of an object is given by the formula \( p = mv \), where \( m \) is the mass and \( v \) is the velocity. For the horizontal component (x-component), \( v_x = 20 \text{ m/s} \). Thus, the initial momentum in the x-direction is \( p_x = 0.4 \text{ kg} \times 20 \text{ m/s} = 8 \text{ kg m/s} \).

Step 2: Determine the Initial Vertical Momentum

When the stone is first thrown, it has no vertical motion because it is thrown horizontally. Therefore, the initial momentum in the y-direction, \( p_y = 0 \).

Step 3: Calculate Time to Hit the Ground

Using the formula for free fall, \( h = \frac{1}{2}gt^2 \), where \( h = 40 \text{ m} \) and \( g = 9.8 \text{ m/s}^2 \), we solve for \( t \): \[ 40 = \frac{1}{2} \times 9.8 \times t^2 \Rightarrow t = \sqrt{\frac{80}{9.8}} \approx 2.857 \text{ s}. \]

Step 4: Determine the Final Vertical Velocity

To find the speed just before hitting the ground, we use the formula \( v_y = gt \). Thus, \( v_y = 9.8 \times 2.857 \approx 28.00 \text{ m/s} \).

Step 5: Calculate Final Momentum Components

At the moment just before impact, the horizontal component of velocity remains constant: \( v_x = 20 \text{ m/s} \). So, \( p_x = 0.4 \text{ kg} \times 20 \text{ m/s} = 8 \text{ kg m/s} \). The vertical component of momentum is \( p_y = 0.4 \text{ kg} \times 28.00 \text{ m/s} = 11.2 \text{ kg m/s} \).

Step 6: Calculate the Impulse Imparted by Gravity

Impulse is the change in momentum. The initial vertical momentum was 0 and just before impact, it became \( 11.2 \text{ kg m/s} \). Thus, the impulse imparted by gravity is \( 11.2 \text{ kg m/s} \).

Key Concepts

Horizontal and Vertical Components

When dealing with momentum, it's crucial to break it down into horizontal and vertical components, especially in projectile motion scenarios. Momentum, represented by \( p \), is calculated as the product of mass \( m \) and velocity \( v \).
For horizontal components, sometimes denoted as the \( x \)-direction, the momentum is calculated using the horizontal velocity. In our example, a stone is thrown horizontally at \( 20 \text{ m/s} \) with a mass of \( 0.4 \text{ kg} \). This provides a horizontal momentum \( p_x \) of \( 8 \text{ kg m/s} \).
Vertical components behave differently. Initially, there is no vertical velocity if the object is thrown purely horizontally. Thus, the initial vertical momentum \( p_y \) is \( 0 \), which will change due to gravitational influences as the projectile gains speed downward.
Pay attention to how separate calculations for each component help in fully understanding the projectile's motion.

Impulse and Change in Momentum

Impulse, in physics, is closely related to momentum. It's the effect of a force acting over time, which results in a change in momentum. Represented by \( J \), it's calculated by the difference in momentum, \( J = \Delta p \).
In our scenario with the stone, the initial vertical momentum \( p_y \) was \( 0 \), as it was just thrown horizontally. By the time the stone was about to hit the ground, gravity increased its vertical speed to \( 28 \text{ m/s} \), resulting in a vertical momentum of \( 11.2 \text{ kg m/s} \). This increase is due to the force of gravity acting over time, which imparts an impulse of \( 11.2 \text{ kg m/s} \).
This demonstrates how impulse accounts for the momentum change, a vital concept in understanding forces in motion.

Projectile Motion

Projectile motion occurs when an object is thrown into the air with two separate velocity components: horizontal and vertical. The horizontal velocity remains constant if we neglect air resistance, while vertical velocity is influenced by gravitational acceleration.
Initially, our stone is only given horizontal velocity \( v_x = 20 \text{ m/s} \). As it travels, gravity starts pulling it downwards, inducing vertical acceleration. This creates a curved path that resembles a parabola.
Understanding projectile motion helps us predict where and when the stone will hit the ground, showing how initial velocities in both directions shape the object's trajectory. Calculating each separately is crucial for analyzing motion accurately.

Free Fall Motion

Free fall motion describes how objects move solely under the influence of gravity. In absence of air resistance, all objects fall at the same rate, regardless of their mass.
For the stone, free fall motion governs its vertical descent. The formula \( h = \frac{1}{2}gt^2 \) helps determine the time \( t \) it takes to hit the ground from a height \( h \). Solving it gave us \( t \approx 2.857 \text{ s} \) for a \( 40 \text{ m} \) drop.
This concept also aids in computing final velocities and momentum changes resulting from gravity. Correctly applying free fall principles allows for precise predictions in vertical motion dynamics, crucial in overall projectile analysis.