Question

A steel ball with mass 40.0 \(g\) is dropped from a height of 2.00 m onto a horizontal steel slab. The ball rebounds to a height of 1.60 m. (a) Calculate the impulse delivered to the ball during impact. (b) If the ball is in contact with the slab for 2.00 ms, find the average force on the ball during impact.

Short answer

(a) Impulse = 0.4744 kg⋅m/s; (b) Average force = 237.2 N.

Step-by-step solution

Understand the Problem

We need to find the impulse delivered to the ball during its impact with the slab and the average force exerted on it. The ball falls from a height 2.00 m, rebounds to a height 1.60 m, with its mass given as 40.0 g, and the time of contact with the slab is 2.00 ms.

Convert Units

The mass of the ball is given as 40.0 g. We need to convert it into kilograms for our calculations: \( m = 40.0 \: g = 0.040 \: kg \).The time of contact is given as 2.00 ms, which can be converted to seconds: \( 2.00 \: ms = 0.002 \: s \).

Calculate Initial and Final Velocities

For the ball dropped from a height, use the energy conservation principle. The velocity just before impact, \( v_i \), can be computed as:\[ v_i = \sqrt{2gh_i} \quad \text{where } g = 9.8 \: m/s^2, \: h_i = 2.00 \: m. \]Plug in the values: \( v_i = \sqrt{2 \times 9.8 \times 2.00} \approx 6.26 \: m/s \).For the ball rebounding to height \( h_f = 1.60 \: m \), the velocity just after leaving the surface, \( v_f \), is:\[ v_f = \sqrt{2gh_f} \]Plug in the values: \( v_f = \sqrt{2 \times 9.8 \times 1.60} \approx 5.60 \: m/s \).

Calculate Impulse

Impulse is defined as the change in momentum: \( J = \Delta p = m(v_f - (-v_i)) \). Substitute \( m = 0.040 \: kg, \: v_f = 5.60 \: m/s, \: v_i = 6.26 \: m/s \) into:\[ J = 0.040 (5.60 + 6.26) = 0.040 \times 11.86 = 0.4744 \: kg \cdot m/s \].

Calculate Average Force

The average force during impact \( F_{avg} \) can be calculated using the formula: \[ F_{avg} = \frac{J}{\Delta t} \]where \( J = 0.4744 \: kg \cdot m/s \) and \( \Delta t = 0.002 \: s \):\[ F_{avg} = \frac{0.4744}{0.002} = 237.2 \: N \].

Key concepts

Kinetic Energy

Kinetic Energy plays a pivotal role in understanding how objects move and interact after being dropped from a height. When the steel ball is released from a height of 2.00 meters, it is initially at rest. Its potential energy is converted into kinetic energy as it falls towards the ground. Kinetic energy

• depends on mass and velocity
• is given by the equation: \( KE = \frac{1}{2}mv^2 \)
• is a scalar quantity

As the ball falls, its velocity increases due to gravity, and so does its kinetic energy. At the moment right before it strikes the slab, all its potential energy has been transformed into kinetic energy.
Kinetic energy at impact can be calculated using the velocity obtained just before the ball hits the slab, which is 6.26 m/s here. This impact velocity, along with the ball's mass, determines the maximum kinetic energy it carries upon striking the solid surface. Thus, understanding kinetic energy is crucial for analyzing the impact phenomena and the rebound height calculations.

Conservation of Energy

The principle of Conservation of Energy is essential for solving this exercise, especially in finding the velocities before and after the impact of the steel ball.Initially, when the ball is at a height of 2.00 meters and 1.60 meters respectively, energy is conserved between potential and kinetic forms.

• Before impact, all gravitational potential energy is converted into kinetic energy.
• Upon impact and rebound, kinetic energy transformed back into potential energy as the ball rises.

This conversion is governed by the equation:\[ mgh = \frac{1}{2}mv^2 \]where \( h \) is the height, \( m \) is mass, \( g \) is gravity, and \( v \) is velocity.
Applying this principle enabled us to calculate the initial and final velocities, 6.26 m/s and 5.60 m/s respectively, by associating the original heights with potential energy. Understanding these energy transformations is key to explain how the ball rebounces and the energy dissipates during the impact with the slab.

Average Force Calculation

Average Force Calculation is a straightforward yet vital aspect of mechanics when finding the force applied over a short timeframe. In this scenario, the average force during the ball's impact with the slab is calculated.With the time of contact given as 2 milliseconds (ms), this duration when multiplied by the rate of change of momentum gives the average force. The formula used is:\[ F_{avg} = \frac{J}{ riangle t} \]where \( J \) represents the impulse or change in momentum, and \( \triangle t \) is the time duration (0.002 seconds)Substituting in the calculated impulse of 0.4744 kg·m/s leads to:\[ F_{avg} = \frac{0.4744}{0.002} \approx 237.2 \text{ N} \]
This calculation shows the force experienced by the ball during its contact with the slab. Average force is important to understand how collisions affect objects, which in turn helps in designing materials and systems that can withstand high forces or impacts efficiently.