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Two ice skaters, Daniel (mass 65.0 \(\mathrm{kg}\) ) and Rebeca (mass 45.0 \(\mathrm{kg}\) , are practicing. Daniel stops to tie his shoelace and, while at rest, is struck by Rebecca, who is moving at 13.0 \(\mathrm{m} / \mathrm{s}\) before she collides with him. After the collision, Rebecca has a velocity of magnitude 8.00 \(\mathrm{m} / \mathrm{s}\) at an angle of \(53.1^{\circ} \mathrm{from}\) her initial direction. Both skaters move on the frictionless, horizontal surface of the rink. (a) What are the magnitude and direction of Daniel's velocity after the collision? (b) What is the change in total kinetic energy of the two skaters as a result of the collision?

Short Answer

Expert verified
Daniel's velocity is approximately 4.02 m/s at an angle of 37° below the x-axis. The change in kinetic energy is -565.5 J.

Step by step solution

01

Analyze the Problem

We need to use the principle of conservation of momentum since there are no external forces. For the x (horizontal) and y (vertical) components, the sum of the initial momenta must equal the sum of the final momenta.
02

Initial Momentum

Write the initial momentum expressions:- Initial momentum of Daniel: Since Daniel is at rest, his momentum is 0.- Initial momentum of Rebecca: Since she moves only in the x-direction initially, it is:\[ p_{x,i} = m_R \cdot v_R = 45.0 \cdot 13.0 = 585.0 \, \mathrm{kg \cdot m/s} \]
03

Final Momentum in X-Direction

After the collision:- Rebecca's x-component of momentum:\[ p_{Rx,f} = m_R \cdot v_R \cdot \cos(53.1^\circ) = 45.0 \cdot 8.0 \cdot \cos(53.1^\circ) \]Calculate this to find \( p_{Rx,f} \). The angle is measured from the x-axis.
04

Final Momentum in Y-Direction

- Rebecca's y-component of momentum:\[ p_{Ry,f} = m_R \cdot v_R \cdot \sin(53.1^\circ) = 45.0 \cdot 8.0 \cdot \sin(53.1^\circ) \]Calculate this to find \( p_{Ry,f} \).
05

Conservation of Momentum Equations

For the x-direction, apply conservation of momentum:\[ 585.0 = p_{Dx, f} + p_{Rx, f} \]For the y-direction:\[ 0 = p_{Dy, f} + p_{Ry, f} \]
06

Solve for Daniel's Momentum Components

Calculate Daniel's x and y components of momentum:\[ p_{Dx, f} = 585.0 - p_{Rx, f} \]\[ p_{Dy, f} = -p_{Ry, f} \]
07

Calculate Daniel's Velocity Magnitude and Direction

Use these components to find the magnitude and direction of Daniel's velocity:\[ v_D = \sqrt{\left(\frac{p_{Dx, f}}{m_D}\right)^2 + \left(\frac{p_{Dy, f}}{m_D}\right)^2} \]Direction (angle with x-axis) is given by:\[ \theta_D = \tan^{-1}\left(\frac{p_{Dy, f}}{p_{Dx, f}}\right) \]
08

Calculate Initial and Final Kinetic Energy

Find initial and final kinetic energies to compute the change:- Initial KE:\[ KE_i = \frac{1}{2} m_R v_R^2 = \frac{1}{2} \times 45.0 \times 13.0^2 \]- Final KE of both:\[ KE_f = \frac{1}{2} m_R \times 8.0^2 + \frac{1}{2} m_D v_D^2 \]
09

Compute the Change in Total Kinetic Energy

The change in kinetic energy is:\[ \Delta KE = KE_f - KE_i \]

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Key Concepts

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

collision physics
In collision physics, understanding how objects interact when they come into contact is fundamental. A collision involves two or more objects exerting forces on each other in a relatively short period. In this scenario, Daniel and Rebecca, two ice skaters, experience such an interaction. In physics, collisions are categorized into two main types: elastic and inelastic. An elastic collision is one in which both momentum and kinetic energy are conserved. In contrast, an inelastic collision conserves momentum but not kinetic energy. This distinction is crucial as it affects both energy distribution and post-collision velocities. For the ice skaters, analyzing the collision helps determine post-impact conditions like speed and direction. The absence of friction (thanks to the ice surface) simplifies the conservation equations, allowing us to focus solely on momentum and kinetic energy changes.
kinetic energy change
The change in kinetic energy during a collision provides insights into what type of collision occurred and how energy is redistributed. Kinetic energy is given by the formula: \[ KE = \frac{1}{2}mv^2 \]where \(m\) is the mass and \(v\) is the velocity of the object.Initial kinetic energy is solely from Rebecca since Daniel is at rest. Post-collision, both skaters move, sharing the kinetic energy. The change in kinetic energy, \( \Delta KE \), can show whether mechanical energy is dissipated (as heat, sound, etc.), which is common in inelastic collisions. Calculating \( \Delta KE \) involves evaluating kinetic energies before and after the collision and finding the difference. If \( \Delta KE \) is significant, it's a sign of a largely inelastic collision, indicating considerable energy distribution in forms other than motion.
principles of momentum
The principles of momentum form the backbone of predicting collision outcomes. Momentum, a conserved quantity in isolated systems, is defined as: \[ p = mv \]where \(p\) is momentum, \(m\) is mass, and \(v\) is velocity. Conservation of momentum states that in the absence of external forces, the total momentum before a collision is equal to the total momentum after. In our example of Daniel and Rebecca, the momentum conservation principle is applied in the x and y directions separately.- Before collision: Only Rebecca contributes to initial horizontal (x) momentum, as Daniel is still.- After collision: Both skaters move with new velocities, maintaining total initial momentum.This principle helps us derive equations to solve for unknowns like Daniel’s speed and direction post-collision, revealing how momentum distributes between involved objects.
frictionless surface dynamics
A frictionless surface, like ice, provides an ideal scenario to study isolated dynamics, particularly useful for conceptual clarity in physics problems. On a frictionless surface, there's negligible resistance to an object’s motion once force is applied, allowing momentum and energy principles to be the main factors impacting motion changes. For ice skaters, this means external forces (like friction or drag) don't interfere, purely localizing the study of interaction to the collision itself. This frictionless interaction simplifies calculations by negating factors like energy losses due to friction. Such simplification highlights momentum conservation and energy redistribution more clearly, making it a valuable scenario for theoretical problem-solving. Emphasizing how forces and motion behave without friction's influence amplifies understanding of underlying physical laws governing motion.

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Most popular questions from this chapter

In a shipping company distribution center, an open cart of mass 50.0 \(\mathrm{kg}\) is rolling to the left at a speed of 5.00 \(\mathrm{m} / \mathrm{s}(\text { Fig } .8 .46)\) . You can ignore friction between the cart and the floor. A 15.0 \(\mathrm{kg}\) package slides down a chute that is inclined at \(37^{\circ}\) from the horizontal and leaves the end of the chute with a speed of 3.00 \(\mathrm{m} / \mathrm{s}\) . The package lands in the cart and they roll off together. If the lower end of the chute is a vertical distance of 4.00 \(\mathrm{m}\) above the bottom of the cart, what are (a) the speed of the package just before it lands in the cart and (b) the final speed of the cart?

An \(8.00-\mathrm{kg}\) ball, hanging from the ceiling by a light wire 135 \(\mathrm{cm}\) long, is struck in an elastic collision by a \(2.00-\mathrm{kg}\) ball moving horizontally at 5.00 \(\mathrm{m} / \mathrm{s}\) just before the collision. Find the tension in the wire just after the collision.

Suppose you hold a small ball in contact with, and directly over, the center of a large ball. If you then drop the small ball a short time after dropping the large ball, the small ball rebounds with surprising speed. To show the extreme case, ignore air resistance and suppose the large ball makes an clastic collision with the floor and then rebounds to make an elastic collision with the still- descending small ball. Just before the collision between the two balls, the large ball is moving upward with velocity \(\overrightarrow{\boldsymbol{v}}\) and the small ball has velocity \(-\overrightarrow{\boldsymbol{v}}\) . (Do you see why? Assume the large ball has a much greater mass than the small ball. (a) What is the velocity of the small ball immediately after its collision with the large hall? (b) From the answer to part (a), what is the ratio of the small ball's rebound distance to the distance it fell before the collision?

Just before it is struck by a racket, a tennis ball weighing 0.560 \(\mathrm{N}\) has a velocity of \((20.0 \mathrm{m} / \mathrm{s}) \hat{\imath}-(4.0 \mathrm{m} / \mathrm{s}) \hat{\mathrm{J}}\) . During the 3.00 \(\mathrm{ms}\) that the racket and ball are in contact, the net force on the ball is constant and equal to \(-(380 \mathrm{N}) \hat{\imath}+(110 \mathrm{N}) \mathrm{J}\) . (a) What are the \(x\) -and \(y\) components of the impulse of the net force applied to the ball? \((b)\) What are the \(x-\) and \(y\) -components of the final velocity of the ball?

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