Question

Starting at rest, two students stand on 10.0 -kg sleds, which point away from each other on ice, and they pass a 5.00 -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?

Short answer

a) Velocity of the student on the left after the throw: -1.0 m/s (moving to the left) b) Velocity of the student on the right after catching the ball: -1.11 m/s (moving to the left) c) Velocity of the student on the left after catching the ball from the right: -1.2 m/s (moving to the left) d) Velocity of the student on the right after the pass: -1.33 m/s (moving to the left)

Step-by-step solution

Calculate initial total momentum

Since the students are at rest, the initial total momentum is zero.

Calculate momentum after the throw

After the throw, the student on the left has thrown the ball with a relative speed of 10.0 m/s to the right. Momentum after the throw = momentum of student + momentum of the ball Since the initial momentum is zero, we get: \(\mathbf{momentum\,of\,student\,on\,the\,left} = -\,\mathbf{momentum\,of\,the\,ball}\)

Calculate the student's velocity after the throw

Using the momentum equation: momentum = mass * velocity, we can calculate the student's velocity. \(\mathbf{velocity\,of\,student\,on\,the\,left} = \frac{-\,\mathbf{momentum\,of\,the\,ball}}{\mathbf{mass\,of\,student\,on\,the\,left}}\) The mass of the student on the left is 50.0 kg, and mass of the ball is 5.00 kg. The relative speed of the ball is 10.0 m/s. \(\mathbf{velocity\,of\,student\,on\,the\,left} = \frac{-(5.00\,\text{kg})(10.0\,\text{m/s})}{50.0\,\text{kg}} = \boxed{-1.0\,\text{m/s}}\) The student on the left is moving at 1.0 m/s to the left (the negative sign indicates direction) after the throw. #b) Velocity of the student on the right after catching the ball#

Calculate momentum after the catch

Momentum after the catch = momentum of student + momentum of the ball Again, since the initial momentum is zero, we get: \(\mathbf{momentum\,of\,student\,on\,the\,right} = -\,\mathbf{momentum\,of\,the\,ball}\)

Calculate the student's velocity after the catch

Using the momentum equation: \(\mathbf{velocity\,of\,student\,on\,the\,right} = \frac{-\,\mathbf{momentum\,of\,the\,ball}}{\mathbf{mass\,of\,student\,on\,the\,right}}\) The mass of the student on the right is 45.0 kg. \(\mathbf{velocity\,of\,student\,on\,the\,right} = \frac{-(5.00\,\text{kg})(10.0\,\text{m/s})}{45.0\,\text{kg}} = \boxed{-1.11\,\text{m/s}}\) The student on the right is moving at 1.11 m/s to the left (negative sign indicates direction) after catching the ball. #c) Velocity of the student on the left after catching the ball from the right#

Calculate momentum after the pass

Momentum after the student on the right passes the ball = momentum of student + momentum of the ball Since the initial momentum is zero, we get: \(\mathbf{momentum\,of\,student\,on\,the\,left} = -\,\mathbf{momentum\,of\,the\,ball}\) \(\mathbf{velocity\,of\,student\,on\,the\,left_AFTER SECOND\_CATCH} = \frac{-\,\mathbf{momentum\,of\,the\,ball}}{\mathbf{mass\,of\,student\,on\,the\,left}}\) The student on the right throws the ball with a relative speed of 12.0 m/s. \(\mathbf{velocity\,of\,student\,on\,the\,left_AFTER SECOND\_CATCH} = \frac{-(5.00\,\text{kg})(12.0\,\text{m/s})}{50.0\,\text{kg}} = \boxed{-1.2\,\text{m/s}}\) After catching the pass from the student on the right, the student on the left is moving at 1.2 m/s to the left (negative sign indicates direction). #d) Velocity of the student on the right after the pass#

Calculate momentum after the pass

Momentum after the student on the right passes the ball = momentum of student + momentum of the ball Since the initial momentum is zero, we get: \(\mathbf{momentum\,of\,student\,on\,the\,right_AFTER SECOND\_THROW} = -\,\mathbf{momentum\,of\,the\,ball}\)

Calculate the student's velocity after the pass

\(\mathbf{velocity\,of\,student\,on\,the\,right_AFTER SECOND\_THROW} = \frac{-\,\mathbf{momentum\,of\,the\,ball}}{\mathbf{mass\,of\,student\,on\,the\,right}}\) \(\mathbf{velocity\,of\,student\,on\,the\,right_AFTER SECOND\_THROW} = \frac{-(5.00\,\text{kg})(12.0\,\text{m/s})}{45.0\,\text{kg}} = \boxed{-1.33\,\text{m/s}}\) The student on the right is moving at 1.33 m/s to the left (negative sign indicates direction) after passing the ball back to the student on the left.

Key concepts

Momentum Calculation

Understanding momentum calculation is crucial for analyzing the movement of objects after interactions like collisions or throws. Momentum, a measure of an object's motion, is calculated by the product of its mass and velocity, expressed by the equation \( p = m \cdot v \).

In the provided exercise, we begin by considering the system's initial state, where both students and sleds are at rest on ice. 'At rest' means their initial velocities, and hence their momenta, are zero. When the student on the left throws the medicine ball, we apply momentum calculation to determine the resulting velocities of the students. Since momentum is a vector quantity, possessing both magnitude and direction, the negative sign obtained in our calculation represents motion in the opposite direction to that of the thrown ball.

The exercise effectively illustrates Newton's third law of motion, which states that for every action, there is an equal and opposite reaction. When applying this to momentum calculations, throwing the ball to the right results in the student moving left with a momentum equal in magnitude but opposite in direction, ensuring total momentum remains conserved.

Relative Speed

Relative speed plays a significant role when examining how objects interact with each other, such as in the passing of the medicine ball between the students on ice.

The term 'relative speed' refers to the velocity of an object in relation to another. In our exercise, the ball's speed is given relative to the student who throws it, not to the ice or any other stationary reference point. It is essential to note that relative speed affects how an object's momentum changes. For instance, when the student on the right catches the ball thrown at 10.0 m/s, this relative speed is crucial for calculating the resulting change in momentum and hence the student's velocity.

An interesting observation from this activity is determining the relative speed of the ball when thrown back by the student on the right at 12.0 m/s. The speed is not simply additive but rather changes the momentum of the system by a different amount due to differences in relative velocity, reflecting how relative speed affects the interaction outcomes between objects.

Conservation of Momentum in Collisions

The principle of conservation of momentum is a foundational concept in physics, particularly during collisions or any interactive events between objects. It states that, in the absence of external forces, the total momentum of a system remains constant.

As seen in the exercise, when the two students pass a medicine ball back and forth, the momentum before and after each throw must remain zero, given that there are no external forces (since friction and air resistance are negligible). This stipulation is reflected in each step of the solution, where the momentum of the student is calculated to be equal in magnitude and opposite in direction to that of the ball after each throw and catch.

Demonstrating the conservation of momentum in a real-world scenario, like the one described in our textbook problem, solidifies students' understanding of the concept. Remember that the conservation doesn't just apply to linear motion - it's also relevant in rotational dynamics, where the conservation is about angular momentum instead. It's these universal principles that bridge the gap between simple textbook problems and the complex movement patterns we observe in the universe around us.