9.3 Conservation of linear momentum (Page 5/10)

University physics volume 1 Page 5 / 10

Significance

It is important to realize that the answer to part (c) is not a velocity; it is a change of velocity, which is a very different thing. Nevertheless, to give you a feel for just how small that change of velocity is, suppose you were moving with a velocity of

4.7 \times 10^{−25} m/s

. At this speed, it would take you about 7 million years to travel a distance equal to the diameter of a hydrogen atom.

Check Your Understanding Would the ball’s change of momentum have been larger, smaller, or the same, if it had collided with the floor and stopped (without bouncing)?

Would the ball’s change of momentum have been larger, smaller, or the same, if it had collided with the floor and stopped (without bouncing)?

If the ball does not bounce, its final momentum ${\vec{p}}_{2}$ is zero, so
$\begin{matrix} Δ \vec{p} & = {\vec{p}}_{2} - {\vec{p}}_{1} \\ = (0) \hat{j} - (−1.4 kg \cdot m/s) \hat{j} \\ = + (1.4 kg \cdot m/s) \hat{j} \end{matrix}$

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Ice hockey 1

Two hockey pucks of identical mass are on a flat, horizontal ice hockey rink. The red puck is motionless; the blue puck is moving at 2.5 m/s to the left ( [link] ). It collides with the motionless red puck. The pucks have a mass of 15 g. After the collision, the red puck is moving at 2.5 m/s, to the left. What is the final velocity of the blue puck?

Two hockey pucks are shown. The top diagram shows the puck on the left with 0 meters per second and the puck on the right moving to the left with 2.5 meters per second. The bottom diagram shows the puck on the left moving to the left at 2.5 meters per second and the puck on the right moving with unknown v. — Two identical hockey pucks colliding. The top diagram shows the pucks the instant before the collision, and the bottom diagram show the pucks the instant after the collision. The net external force is zero.

Strategy

We’re told that we have two colliding objects, we’re told the masses and initial velocities, and one final velocity; we’re asked for both final velocities. Conservation of momentum seems like a good strategy. Define the system to be the two pucks; there’s no friction, so we have a closed system.

Before you look at the solution, what do you think the answer will be?

The blue puck final velocity will be:

zero
2.5 m/s to the left
2.5 m/s to the right
1.25 m/s to the left
1.25 m/s to the right
something else

Solution

Define the + x -direction to point to the right. Conservation of momentum then reads

\begin{matrix} {\vec{p}}_{f} & = & {\vec{p}}_{i} \\ m v_{r_{f}} \hat{i} + m v_{b_{f}} \hat{i} & = & m v_{r_{i}} \hat{i} - m v_{b_{i}} \hat{i} . \end{matrix}

Before the collision, the momentum of the system is entirely and only in the blue puck. Thus,

\begin{matrix} m v_{r_{f}} \hat{i} + m v_{b_{f}} \hat{i} & = & − m v_{b_{i}} \hat{i} \\ v_{r_{f}} \hat{i} + v_{b_{f}} \hat{i} & = & − v_{b_{i}} \hat{i} . \end{matrix}

(Remember that the masses of the pucks are equal.) Substituting numbers:

\begin{matrix} - (2.5 m/s) \hat{i} + {\vec{v}}_{b_{f}} & = & − (2.5 m/s) \hat{i} \\ {\vec{v}}_{b_{f}} & = & 0 . \end{matrix}

Significance

Evidently, the two pucks simply exchanged momentum. The blue puck transferred all of its momentum to the red puck. In fact, this is what happens in similar collision where

m_{1} = m_{2} .

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Check Your Understanding Even if there were some friction on the ice, it is still possible to use conservation of momentum to solve this problem, but you would need to impose an additional condition on the problem. What is that additional condition?

Consider the impulse momentum theory, which is $\vec{J} = Δ \vec{p}$ . If $\vec{J} = 0$ , we have the situation described in the example. If a force acts on the system, then $\vec{J} = {\vec{F}}_{ave} Δ t$ . Thus, instead of ${\vec{p}}_{f} = {\vec{p}}_{i}$ , we have
${\vec{F}}_{ave} Δ t = Δ \vec{p} = {\vec{p}}_{f} - {\vec{p}}_{i}$
where ${\vec{F}}_{ave}$ is the force due to friction.

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Landing of Philae

On November 12, 2014, the European Space Agency successfully landed a probe named Philae on Comet 67P/Churyumov/Gerasimenko ( [link] ). During the landing, however, the probe actually landed three times, because it bounced twice. Let’s calculate how much the comet’s speed changed as a result of the first bounce.

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Source: OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5

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