Question

Starting at rest, two students stand on \(10.0-\mathrm{kg}\) sleds, which point away from each other on ice, and they pass a \(5.00-\mathrm{kg}\) medicine ball back and forth. The student on the left has a mass of \(50.0 \mathrm{~kg}\) and can throw the ball with a relative speed of \(10.0 \mathrm{~m} / \mathrm{s} .\) The student on the right has a mass of \(45.0 \mathrm{~kg}\) and can throw the ball with a relative speed of \(12.0 \mathrm{~m} / \mathrm{s}\). (Assume there is no friction between the ice and the sleds and no air resistance.) a) If the student on the left throws the ball horizontally to the student on the right, how fast is the student on the left moving right after the throw? b) How fast is the student on the right moving right after catching the ball? c) If the student on the right passes the ball back, how fast will the student on the left be moving after catching the pass from the student on the right? (d) How fast is the student on the right moving after the pass? a) If the student on the left throws the ball horizontally to the student on the right, how fast is the student on the left moving right after the throw? b) How fast is the student on the right moving right after catching the ball? c) If the student on the right passes the ball back, how fast will the student on the left be moving after catching the pass from the student on the right? (d) How fast is the student on the right moving after the pass?

Short answer

Question: Two students are on frictionless sleds. The student on the left has a mass of 50.0 kg, and the student on the right has a mass of 45.0 kg. They have a 5.00 kg medicine ball. The student on the left throws the medicine ball at 10.0 m/s horizontally to the student on the right who catches it and throws it back at 12.0 m/s. Calculate the speeds of the students after: a) The student on the left throws the ball. b) The student on the right catches the ball. c) The student on the left catches the ball thrown back by the student on the right. d) The student on the right throws the ball back. Answer: a) 1.00 m/s b) 1.11 m/s c) 2.31 m/s d) 2.58 m/s

Step-by-step solution

(Step 1: Calculate the momentum before throwing)

Before the ball is thrown, the total momentum of the system is zero, as both the students and the ball are at rest.

(Step 2: Calculate the momentum after the first throw and find the student's speed)

To calculate the momentum after the student on the left throws the ball, we'll use the equation p = mv, where p is momentum, m is mass, and v is velocity. Since the total momentum after the throw must remain zero, the momentum of the student on the left after throwing the ball must be equal and opposite to the momentum of the ball. Using the given data: Mass of Student (left) = m1 = 50.0 kg Mass of Ball = m2 = 5.00 kg Relative speed of Ball thrown by Student (left) = v2 = 10.0 m/s. Momentum of Ball after being thrown = p2 = m2 * v2 = 5.00 kg * 10.0 m/s = 50.0 kg·m/s Therefore, Momentum of Student (left) after throwing the ball = -50.0 kg·m/s (opposite direction to Ball) Now, we can find the velocity of the student on the left (v1) after the throw: v1 = -p1 / m1 = - (-50.0 kg·m/s) / 50.0 kg = 1.00 m/s. Therefore, the speed of the student on the left after the throw is 1.00 m/s.

(Step 3: Calculate the momentum after the ball is caught and find the second student's speed)

After the student on the right catches the ball, the momentum of that student plus the momentum of the ball must be equal to the initial momentum of the system (zero). Mass of Student (right) = m3 = 45.0 kg Speed of Student (right) after catching the ball = v3 The momentum of the system after catching the ball: p3 = m3 * v3 + p2 = 0 v3 = -p2 / m3 = -50.0 kg·m/s / 45.0 kg = -1.11 m/s So, the speed of the student on the right after catching the ball is 1.11 m/s.

(Step 4: Calculate the speeds after the ball is thrown back and caught)

We'll repeat the process for when the student on the right throws the ball back. We need to consider that both students are already moving when the throw happens. Relative speed of Ball thrown by Student (right) = v4 = 12.0 m/s Momentum of Ball after being thrown back = p4 = m2 * (v4 - v3) = 5.00 kg * (12.0 m/s - (-1.11 m/s)) = 65.6 kg·m/s Now we can find the velocity of the student on the right after passing the ball back to the student on the left: v5 = (m3 * v3 - p4) / m3 = (45.0 kg * (-1.11 m/s) - 65.6 kg·m/s) / 45.0 kg = -2.58 m/s So, the speed of the student on the right after the pass is 2.58 m/s. Finally, we can find the speed of the student on the left after catching the ball: v6 = (m1 * v1 + p4) / m1 = (50.0 kg * 1.00 m/s + 65.6 kg·m/s) / 50.0 kg = 2.31 m/s So, the speed of the student on the left after catching the pass from the student on the right is 2.31 m/s. In summary: a) The speed of the student on the left after the throw is 1.00 m/s. b) The speed of the student on the right after catching the ball is 1.11 m/s. c) The speed of the student on the left after catching the pass from the student on the right is 2.31 m/s. d) The speed of the student on the right after the pass is 2.58 m/s.

Key concepts

Momentum and Impulse

Understanding momentum and impulse is key to analyzing the movement of objects after collisions or force applications. In physics, momentum, symbolized by 'p', is the product of an object's mass (m) and its velocity (v): \(p = m \times v\). It is a vector quantity, meaning it has both magnitude and direction. Since the two students in our example start at rest and then move away from each other after interacting with the medicine ball, we can visualize momentum as a measure of how 'hard' it would be to stop them.

Impulse is related to momentum and is defined as the change in momentum resulting from a force acting over a period of time. The impulse imparted to an object equals the force (F) multiplied by the time interval (\(\Delta t\)) during which the force is applied. The impulse-momentum theorem states that the impulse on an object is equal to the change in its momentum, which can be mathematically expressed as \(F \times \Delta t = \Delta p\).

In the exercise, when a student throws the ball, they exert a force over a certain time, giving the ball an impulse and therefore changing its momentum. The reaction force (Newton's third law) causes an equal and opposite impulse on the student, changing their momentum in the opposite direction. Understanding this interplay of forces and the resulting change in momentum is crucial to solving problems involving motion and interactions.

Conservation of Momentum Examples

The principle of conservation of momentum is best explored through examples. It states that within a closed system, the total momentum remains constant unless acted upon by an external force. This law is perfectly showcased in the scenario of two students on sleds throwing a medicine ball back and forth on a frictionless surface.

At the start, the combined momentum of the students and ball is zero because they are at rest. When one student throws the ball, the momentum of that student becomes equal and opposite to the momentum of the ball to keep the total momentum at zero. As the other student catches the ball, they gain momentum equal to the momentum of the ball, but in the opposite direction of the ball's initial momentum. This push-and-pull sequence continues with each throw and catch, each action demonstrating momentum conservation.

By calculating the momentum before and after each ball exchange and considering the total system's momentum, we can determine the resultant speeds. This reflects real-world applications like the motion of objects in space, where astronauts use tools or throw objects to propel themselves, conserving momentum in a nearly frictionless environment.

Closed System in Physics

A closed system in physics is a system where no mass enters or leaves, and no external forces act on the objects within the system. In our exercise, the students on the sleds and the medicine ball constitute a closed system. The ice is frictionless, and there is no air resistance, ensuring that no external forces like friction interfere with the movements of the students and the ball.

Because the system is closed, we see the implications of Newton’s laws of motion. The third law, which states that for every action there is an equal and opposite reaction, is observable when the students throw the medicine ball. The reaction force makes the students move in the opposite direction to the throw. Since no external forces are present to change the momentum, the system provides a pure demonstration of the conservation of momentum in action, where the total momentum before and after each throw remains zero. This concept of a closed system is crucial in astrophysics and space exploration, where spacecraft maneuvers are often calculated assuming closed systems to conserve fuel and reach the desired trajectory with precision.