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A lumberjack (mass \(=98 \mathrm{~kg}\) ) is standing at rest on one end of a floating log (mass \(=230 \mathrm{~kg}\) ) that is also at rest. The lumberjack runs to the other end of the log, attaining a velocity of \(+3.6 \mathrm{~m} / \mathrm{s}\) relative to the shore, and then hops onto an identical floating log that is initially at rest. Neglect any friction and resistance between the logs and the water. (a) What is the velocity of the first log just before the lumberjack jumps off? (b) Determine the velocity of the second log if the lumberjack comes to rest on it.

Short Answer

Expert verified
(a) The velocity of the first log is -1.534 m/s. (b) The velocity of the second log is 1.075 m/s.

Step by step solution

01

Understand the Conservation of Momentum

Momentum in an isolated system is conserved when no external forces are acting on it. For the system of the lumberjack and the log, since the system is initially at rest, its total momentum is zero. Thus, the momentum before the lumberjack moves must equal the total momentum after he starts running.
02

Calculate the Velocity of the First Log

By conservation of momentum, the initial momentum of the system (zero) equals the final momentum. Let \( v_1 \) be the velocity of the log. The equation for conservation of momentum is:\[0 = m_{ ext{lumberjack}} \cdot v_{ ext{lumberjack}} + m_{ ext{log}} \cdot v_1\]Substitute the given values:\[0 = 98 \times 3.6 + 230 \cdot v_1\]Solving for \( v_1 \):\[230 \cdot v_1 = -352.8 \v_1 = \frac{-352.8}{230} \v_1 = -1.534 ext{ m/s}\]
03

Analyze the Momentum During the Jump to the Second Log

When the lumberjack jumps onto the second log, the total system includes both the second log and the lumberjack. Before the jump, the second log is stationary with no momentum. Post-jump, the momentum must remain constant under the condition of no external forces. Therefore, the momentum lumberjack takes with him is transferred to the second log.
04

Calculate the Velocity of the Second Log

Using the conservation of momentum for the lumberjack and the second log, let \( v_2 \) be the velocity of the second log. Before the lumberjack lands on the second log, he still possesses his velocity. Thus:\[0 = m_{ ext{lumberjack}} \cdot v_{ ext{lumberjack}} + m_{ ext{second log}} \cdot v_2\]After reaching the second log, the lumberjack and it come to rest relative to each other, so they share a common velocity of \( v_2 \). So:\[98 \cdot 3.6 + 230 \cdot 0 = (98 + 230) \cdot v_2\]Substituting the combined mass and solving for \( v_2 \):\[352.8 = 328 \cdot v_2 \v_2 = \frac{352.8}{328} \v_2 = 1.075 ext{ m/s}\]

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Key Concepts

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

Momentum in Isolated Systems
Momentum is a cornerstone concept in physics, integral to understanding motion in isolated systems. The law of conservation of momentum tells us that within such systems, the total momentum remains constant. This means that if no external forces are at play—such as friction or air resistance—any momentum gain by one object must be offset by an equal and opposite momentum change in another.
In our example, the lumberjack and the log floating on the water form an isolated system. They are initially at rest, which implies their total momentum is zero. When the lumberjack starts moving, he creates momentum in the system. However, to satisfy the conservation law, the log must move in the opposite direction to maintain a net momentum of zero.
  • This principle ensures that motion in one part of the system does not disturb the overall balance, making it vital in analyzing scenarios like collisions or separations.
  • Understanding momentum conservation helps in diverse applications from vehicle collisions to motion in space.
Velocity Calculation
Calculating velocity in scenarios involving conservation of momentum involves working through formulaic steps. Starting from the basic momentum equation, we can establish the relationship:
For any object, momentum is the product of its mass and velocity: \[ P = m \cdot v \]
When considering two objects interacting, such as the lumberjack and the log, we set up the conservation equation for momentum:
\[ m_{\text{jack}} \cdot v_{\text{jack}} = -m_{\text{log}} \cdot v_{\text{log}} \]
  • The minus sign indicates the opposite directions of motion.
  • After plugging in known values, solve for the unknown velocity using algebraic manipulation.
In our case with the lumberjack, the calculated negative velocity for the log signifies its movement in the opposite direction relative to the lumberjack. Each calculation is insightful and reaffirms how velocities balance to preserve momentum in isolated systems.
Motion in Physics
Motion in physics, particularly when observing from a conservation standpoint, involves analyzing how objects interact and impact one another. When the lumberjack begins to move, he imparts motion onto the log due to conservation laws.
  • Understanding this interaction requires breaking down individual motions and aggregating them to see the broader picture.
  • This exercise illuminates how even seemingly simple actions, like running across a log, involve complex physics principles.
In the lumberjack's case, when he jumps on and off the logs, he shifts momentum to them, affecting their motion and velocity. It beautifully demonstrates how conservation laws manifest in everyday events, making it tangible and observable for learners.

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Most popular questions from this chapter

A \(0.500-\mathrm{kg}\) ball is dropped from rest at a point \(1.20 \mathrm{~m}\) above the floor. The ball rebounds straight upward to a height of \(0.700 \mathrm{~m}\). What are the magnitude and direction of the impulse of the net force applied to the ball during the collision with the floor?

A golfer, driving a golf ball off the tee, gives the ball a velocity of \(+38 \mathrm{~m} / \mathrm{s}\). The mass of the ball is \(0.045 \mathrm{~kg},\) and the duration of the impact with the golf club is \(3.0 \times 10^{-3} \mathrm{~s}\). (a) What is the change in momentum of the ball? (b) Determine the average force applied to the ball by the club.

When jumping straight down, you can be seriously injured if you land stiff- legged One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. A 75 -kg man just before contact with the ground has a speed of \(6.4 \mathrm{~m} / \mathrm{s}\). (a) In a stiff-legged landing he comes to a halt in \(2.0 \mathrm{~ms}\). Find the average net force that acts on him during this time. (b) When he bends his knees, he comes to a halt in \(0.10 \mathrm{~s}\). Find the average net force now. (c) During the landing, the force of the ground on the man points upward, while the force due to gravity points downward. The average net force acting on the man includes both of these forces. Taking into account the directions of the forces, find the force of the ground on the man in parts (a) and (b).

ssm Batman (mass \(=91 \mathrm{~kg}\) ) jumps straight down from a bridge into a boat (mass \(=510 \mathrm{~kg}\) ) in which a criminal is fleeing. The velocity of the boat is initially \(+11\) \(\mathrm{m} / \mathrm{s}\). What is the velocity of the boat after Batman lands in it?

Two ice skaters have masses \(m_{1}\) and \(m_{2}\) and are initially stationary. Their skates are identical. They push against one another, as in Figure \(7-11,\) and move in opposite directions with different speeds. While they are pushing against each other, any kinetic frictional forces acting on their skates can be ignored. However, once the skaters separate, kinetic frictional forces eventually bring them to a halt. As they glide to a halt, the magnitudes of their accelerations are equal, and skater 1 glides twice as far as skater 2\. What is the ratio \(m_{1} / m_{2}\) of their masses?

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