Problem 93 - University Physics with Modern Physics

Question

A 170.-g hockey puck moving in the positive \(x\) direction at \(30.0 \mathrm{~m} / \mathrm{s}\) is struck by a stick at time \(t=2.00 \mathrm{~s}\) and moves in the opposite direction at \(25.0 \mathrm{~m} / \mathrm{s}\). If the puck is in contact with the stick for \(0.200 \mathrm{~s}\), plot the momentum and the position of the puck, and the force acting on it as a function of time, from 0 to \(5.00 \mathrm{~s}\). Be sure to label the coordinate axes with reasonable numbers.

Step-by-Step Solution

Answer

Question: Plot the momentum, position, and force acting on a hockey puck after being struck by a stick and analyze its motion. Answer: To analyze the motion of the puck, we calculate its initial and final momentum, find the change in momentum, calculate the average force exerted by the stick, and then plot the momentum, position, and force as a function of time. Initial momentum: \(p_i = 5.10 \mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}\) Final momentum: \(p_f = -4.25 \mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}\) Change in momentum: \(\Delta p = -9.35 \mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}\) Average force: \(F = -46.75 \mathrm{~N}\) Momentum as a function of time: - \(0 \leq t < 2.00\mathrm{~s}\): \(5.10\mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}\) - \(2.00\mathrm{~s} \leq t \leq 2.20\mathrm{~s}\): linear change from \(5.10\mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}\) to \(-4.25\mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}\) - \(t > 2.20\mathrm{~s}\): \(-4.25\mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}\) Position as a function of time: - \(0 \leq t < 2.00\mathrm{~s}\): \(x(t) = 30.0t\) - \(2.00\mathrm{~s} \leq t \leq 2.20\mathrm{~s}\): \(x(t) = at^2 + bt + c\) - \(t > 2.20\mathrm{~s}\): \(x(t) = -25.0(t-2.20)\) Force as a function of time: - \(0 \leq t < 2.00\mathrm{~s}\): \(F = 0\mathrm{~N}\) - \(2.00\mathrm{~s} \leq t \leq 2.20\mathrm{~s}\): \(F = -46.75\mathrm{~N}\) - \(t > 2.20\mathrm{~s}\): \(F = 0\mathrm{~N}\)

Step 1: Calculate Initial Momentum

To find the initial momentum of the puck, we can use the formula for momentum, which is \(p = m \cdot v\), where \(p\) is the momentum, \(m\) is the mass, and \(v\) is the velocity. For the initial velocity, we are given \(30.0 \mathrm{~m}/\mathrm{s}\) in the positive x direction. The mass of the puck is \(170 \mathrm{g}\), which we need to convert to kilograms: \(m = 170 \cdot 10^{-3} \mathrm{~kg}=0.170 \mathrm{~kg}\). The initial momentum is: $$ p_{i} = (0.170 \mathrm{~kg})(30.0 \mathrm{~m}/\mathrm{s}) = 5.10 \mathrm{~kg} \cdot \mathrm{m}/\mathrm{s} $$

Step 2: Calculate Final Momentum

After being hit, the puck moves in the opposite direction at \(-25.0 \mathrm{~m}/\mathrm{s}\). The mass of the puck remains the same, so the final momentum is: $$ p_{f} = (0.170 \mathrm{~kg})(-25.0 \mathrm{~m}/\mathrm{s}) = -4.25 \mathrm{~kg} \cdot \mathrm{m}/\mathrm{s} $$

Step 3: Calculate Change in Momentum and Force

Now we can find the change in momentum, \(\Delta p\), by subtracting the initial momentum \(p_{i}\) from the final momentum \(p_{f}\): $$ \Delta p = p_{f} - p_{i} = -4.25 - 5.10 = -9.35 \mathrm{~kg} \cdot \mathrm{m}/\mathrm{s} $$ The puck is in contact with the stick for \(0.200\mathrm{~s}\), so to find the average force exerted by the stick, we can use the formula for impulse, which is \(F \cdot \Delta t = \Delta p\), where \(\Delta p\) is the change in momentum, \(F\) is the force, and \(\Delta t\) is the time interval. So the average force is: $$ F = \frac{\Delta p}{\Delta t} = \frac{-9.35 \mathrm{~kg} \cdot \mathrm{m}/\mathrm{s}}{0.200 \mathrm{~s}} = -46.75 \mathrm{~N} $$

Step 4: Plot Momentum as a Function of Time

To plot the momentum as a function of time, we will have a piecewise function. Before the puck is hit (\(0 \leq t < 2.00\mathrm{~s}\)), its momentum remains constant at \(5.10\mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}\). Between \(2.00\mathrm{~s}\) and \(2.20\mathrm{s}\) (during the contact with stick), the momentum changes linearly from \(5.10\mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}\) to \(-4.25\mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}\). After the puck is hit (\(t > 2.20\mathrm{~s}\)), its momentum remains constant at \(-4.25\mathrm{~kg}\cdot\mathrm{m}/\mathrm{s}\).

Step 5: Plot Position as a Function of Time

To plot the position as a function of time, we can integrate the velocity with respect to time. Before the puck is hit (\(0 \leq t < 2.00\mathrm{~s}\)), \(v = 30.0\mathrm{~m}/\mathrm{s}\). So, the position function is \(x(t) = 30.0t\). Between \(2.00\mathrm{~s}\) and \(2.20\mathrm{~s}\) (during the contact with stick), the velocity changes linearly from \(30.0\mathrm{~m}/\mathrm{s}\) to \(-25.0\mathrm{~m}/\mathrm{s}\), so we can assume the position function to be of the form \(x(t) = at^{2} + bt + c\). After the puck is hit (\(t > 2.20\mathrm{~s}\)), \(v = -25.0\mathrm{m}/\mathrm{s}\). So, the position function is \(x(t) = -25.0(t-2.20)\).

Step 6: Plot Force as a Function of Time

To plot the force as a function of time, we will have a piecewise function. Before the puck is hit (\(0 \leq t < 2.00\mathrm{~s}\)), there is no force acting on the puck, so \(F = 0\mathrm{~N}\). Between \(2.00\mathrm{~s}\) and \(2.20\mathrm{~s}\) (during the contact with the stick), the force is constant at \(-46.75\mathrm{~N}\). After the puck is hit (\(t > 2.20\mathrm{~s}\)), there is no force acting on the puck, so \(F = 0\mathrm{~N}\). Now, create a labeled plot of the momentum, position, and force as a function of time, from \(0\mathrm{~s}\) to \(5.00\mathrm{~s}\).

Key Concepts

Impulse and Momentum

To understand the relationship between impulse and momentum, let's start by defining both concepts. Momentum, represented as \(p\), is a measure of the motion of an object and is calculated as the product of an object's mass \(m\) and its velocity \(v\), or \(p = m \times v\). Momentum itself is a vector, indicating that it has both magnitude and direction. Impulse, on the other hand, refers to a change in momentum caused by a force acting over a period of time. \(\Delta p\) or impulse is given by the equation \(\Delta p = F \cdot \Delta t\), where \(F\) is the average force applied and \(\Delta t\) is the time period over which it acts.

Applying this to the given hockey puck scenario, when the puck is struck by the stick, an impulse is applied changing the puck's momentum from its initial state to a final one in the opposite direction. The change in momentum (impulse) can be calculated as the difference between the initial and final momentum vectors. In this example, the impulse can be both positive and negative, which indicates not just the change in magnitude of the momentum but also the reversal in direction. This concept is critical to solving problems involving collisions and interactions between objects where forces are applied over time intervals.

Thus, in the calculation of the force, we see the average force exerted by the stick on the puck is determined by the impulse equation, and the negative force value signifies that the force is applied in the direction opposite to the initial motion of the puck.

Conservation of Momentum

The principle of conservation of momentum states that in a closed system with no external forces, the total momentum remains constant. This principle is widely applicable in physics from macroscopic to atomic scales and plays a crucial role in collision and explosion problems. Within the context of the hockey puck, if we considered the interaction between the puck and stick in isolation and assumed no external forces act (like friction or air resistance), the total momentum of the puck and the stick system before and after the collision would be the same.

In our solved problem, we only focus on the momentum of the puck. However, if we consider the stick as part of the system, the stick's action on the puck would be an internal force, and hence would not change the total momentum of the stick-puck system. It is important to remember, though, that while momentum is conserved, kinetic energy might not be—especially in inelastic collisions. In our case, details regarding the stick's mass and velocity are not mentioned, but if they were, we could apply the conservation of momentum principle to find missing information after considering the forces involved as internal to the system.

Force-Time Graph

A force-time graph visually represents the force applied to an object over a specific time period. The area under the curve on this graph represents the impulse, which is the change in momentum. In the case of the hockey puck, plotting force as a function of time would show a graph where the applied force spikes during the interval the stick and puck are in contact and is zero before and after this event, as no other forces are acting on the puck.

In the step-by-step solution, creating a piecewise function for the force allows us to capture the sudden application and removal of force. On the force-time graph, before the puck is hit (from 0 to 2.00 seconds), the force remains zero, reflecting the fact that no external forces are acting on it. There's a sharp jump to -46.75 N between 2.00 and 2.20 seconds, corresponding to the stick contacting the puck and applying negative force (opposite to the direction of the puck's initial motion), and then it returns to zero after 2.20 seconds as the contact ends. The area under the force-time graph during the contact would then illustrate the impulse applied to the puck, clearly linking the graphical representation with the physical quantity involved.