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A force magnitude that averages \(1200 \mathrm{~N}\) is applied to a \(0.40 \mathrm{~kg}\) steel ball moving at 14 \(\mathrm{m} / \mathrm{s}\) in a collision lasting \(27 \mathrm{~ms}\). If the force is in a direction opposite the initial velocity of the ball, find the final speed and direction of the ball.

Short Answer

Expert verified
Final speed: 81 m/s, direction: opposite initial velocity.

Step by step solution

01

Identify Given Values

Identify all the given values in the problem: - Force magnitude: 1200 N - Mass of the ball: 0.40 kg - Initial velocity: 14 m/s - Collision duration: 27 ms (or 0.027 s after converting milliseconds to seconds)
02

Calculate Impulse

Impulse is calculated using the formula: \[ J = F \times \text{time} \] Substituting the given values: \[ J = 1200 \times 0.027 = 32.4 \text{ Ns} \]
03

Determine Velocity Change

Using the impulse-momentum theorem: \[ J = \text{change in momentum} = m \times (\text{final velocity} - \text{initial velocity}) \] Solving for final velocity: \[ 32.4 = 0.40 \times (v_f - 14) \] Rearranging: \[ v_f - 14 = \frac{32.4}{0.40} = 81 \] Therefore: \[ v_f = 81 + 14 = 95 \text{ m/s} \]
04

Account for Direction

Since the force is in the opposite direction of the initial velocity, the final speed will be in the opposite direction. Thus: \[ v_f = -81 \text{ m/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

force calculation
Understanding force calculation in collision dynamics is crucial. Force, measured in newtons (N), is a vector quantity. This means it has both magnitude and direction. In our example, a force of 1200 N was applied opposite to the ball's initial motion. This is an essential point because it influences the direction of the final velocity. Calculating such a force often involves measuring the impact or using sensors that quantify the applied force during a collision. Accurately knowing this value is the first step to solving the problem, as it directly impacts subsequent steps like impulse and momentum calculations.
impulse
Impulse is a fundamental concept related to force and time. It is defined as the product of the average force applied and the duration of time it is applied for. Its formula is easy: \[ J = F \times \text{time} \] It's measured in Newton-seconds (Ns). In the example, the impulse was calculated using a force of 1200 N over a time span of 27 milliseconds, converting milliseconds to seconds first. This gave us: \[ J = 1200 \times 0.027 = 32.4 \text{ Ns} \] Impulse essentially changes an object's momentum, and it's useful because it simplifies the relation between an applied force and the resulting motion.
momentum
Momentum is another core concept you'll need to grasp. It's a product of an object's mass and velocity, expressed as: \[ p = m \times v \] Measured in kilograms meters per second (kg·m/s). The impulse-momentum theorem connects impulse and momentum, stating that \[ J = \text{Change in momentum} = m \times (\text{final velocity} - \text{initial velocity}) \] This theorem helps us solve for the final velocity. In our exercise, we calculated the change in momentum to determine the final speed of the ball. Plugging in the values: \[ 32.4 = 0.40 \times (v_f - 14) \] we rearranged it to find: \[ v_f = 81 + 14 = 95 \text{ m/s} \] Remember, momentum depends on both the speed and the direction of the moving object.
collision dynamics
Collision dynamics involves understanding the behavior of objects during and after they collide. In our exercise, the force applied was in the opposite direction of the ball's movement. This significantly affects the final outcome. Because the ball was initially moving at 14 m/s, the change in direction was calculated by noting that the final speed was 95 m/s, but in the opposing direction. Therefore: \[ v_f = -81 \text{ m/s} \] When analyzing collisions:
  • Identify forces involved, their magnitude and direction
  • Calculate impulse to understand changes in momentum
  • Use the impulse-momentum theorem to find resulting velocities
Dynamics in collisions also consider whether the collision is elastic or inelastic, which affects how kinetic energy is conserved or converted.

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Most popular questions from this chapter

A bullet of mass \(4.5 \mathrm{~g}\) is fired horizontally into a \(2.4 \mathrm{~kg}\) wooden block at rest on a horizontal surface. The bullet is embedded in the block. The speed of the block immediately after the bullet stops relative to it is \(2.7 \mathrm{~m} / \mathrm{s}\). At what speed is the bullet fired?

In February 1955, a paratrooper fell \(370 \mathrm{~m}\) from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was \(56 \mathrm{~m} / \mathrm{s}\) (terminal speed), that his mass (including gear) was \(85 \mathrm{~kg}\), and that the magnitude of the force on him from the snow was at the survivable limit of \(1.2 \times 10^{5} \mathrm{~N}\). What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?

A billiard ball moving at a speed of \(2.2 \mathrm{~m} / \mathrm{s}\) strikes an identical stationary ball with a glancing blow. After the collision, one ball is found to be moving at a speed of \(1.1 \mathrm{~m} / \mathrm{s}\) in a direction making a \(60^{\circ}\) angle with the original line of motion. Find the velocity of the other ball.

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In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at \(3.50 \mathrm{~m} / \mathrm{s}\) along a line making an angle of \(22.0^{\circ}\) with its original direction of motion, and the second ball has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball.

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