Brock University Physics 1P21/1P91 Fall 2013 Dr. D’Agostino

Solutions for Tutorial 9: Chapter 9, Momentum

The goals of this tutorial are:

• understand momentum, impulse, and the impulse-momentum theorem

• understand angular momentum

• understand a version of the rotational analogue of Newton’s second law of motion,

τ =∆L

∆t

1. You probably have engaged in picnic games such as catching raw eggs or water balloons.If not, you’ve probably engaged in sports involving throwing and catching balls. Brieflyexplain how you would catch a ball (such as a water balloon or a hard-thrown football)in order to minimize the force that the ball exerts on your hands. You do this becauseyou don’t want the balloon to break, or you don’t want the hard-thrown football tohurt your hands.

Include in your answer the action you take, and a physical explanation for your action,quoting a theorem or principle or equation that is relevant. [2 points]

Solution: What we instinctively do is to gradually decelerate the ball over as long atime as possible, by meeting the ball well in front of us, with our arms outsretched, andthen we gradually bring our hands back towards our body (and even past our body,perhaps).

We have no control over the necessary change in momentum needed to stop the ball,because we have no control over the initial momentum of the incoming ball. By theimpulse-momentum theorem, the impulse needed to stop the ball is also not in ourcontrol; it is equal in magnitude to the momentum of the incoming ball.

But the impulse is the product of the average force and the time interval during whichthe force acts. Thus, although we have no control over the product Favg∆t, which isequal in magnitude to the change in momentum, we can reduce Favg by increasing ∆t.

Cars have “crumple zones” in their front ends for the same reason; by increasing thetime over which a car is decelerated during a crash, the average force is decreased, andthis helps protect occupants of the car.

2. As the Earth warms, the polar ice caps melt. Water thereby moves from the ice capsand spreads out around the surface of the oceans. The length of Earth’s days willtherefore change. State whether the day will get longer or shorter, and briefly explainthe change using concepts you’ve learned in this chapter. [2 points]

Solution: Ice in the ice caps is close to the rotation axis, and so contributes a relativelysmall amount to the moment of inertia of the Earth. However, when the ice melts, theresulting water spreads over the surface of the oceans, and so some of the water ends

up close to the equator, where it is much farther from the rotation axis, and thereforecontributes much more to the moment of inertia of the Earth.

By the principle of conservation of angular momentum, the earth’s angular momentumwill remain constant as the ice melts. However, if the moment of inertia increases,then the angular velocity must decrease in order for the angular momentum to remainconstant, because L = Iω. If the angular speed of Earth’s rotation decrases, the lengthof its day will increase.

3. While standing still on a basketball court, you throw a basketball to a teammate. Whydo you not move backward as a result? Is the principle of conservation of momentumviolated? Briefly explain. [2 points]

Solution: The momentum of a system is conserved provided that the net externalforce acting on the system is zero. Let’s consider you and the ball as the system; ifthere were no other forces acting on this system, then yes indeed, you would “recoil”when you pass the ball to your teammate. 1 Consider astronauts in a space station,or outside a space station. Similarly, if you are in a canoe and you throw something tosomeone outside the canoe, then you and your canoe will move a little in the oppositedirection.

When you pass the ball, notice that you “brace” yourself against the floor. In otherwords, there is an external force acting on the system, at least on you, and you are partof the system. The force of the floor on you is what provides the change in momentumof the system that is you + basketball.

Here is another perspective: If you consider the system to be you, the basketball,your teammate, and the Earth, then there are no external forces, and momentum isconserved. When you throw the basketball, increasing its momentum in one direction,you + Earth recoil a bit, and so the momentum of you + Earth increase in the otherdirection; thus, the net change in momentum is zero, and momentum is conserved.However, because the Earth is so massive, its change in velocity is extremely tiny.

When your teammate catches the ball, the ball’s momentum is transferred to yourteammate + Earth (they are attached), so momentum is again conserved. The netchange in momentum of the Earth after the whole process is completed is zero. Simi-larly, the net change in momentum of the ball, of you, and of your teammate, are allalso zero once the whole process is complete.

So, no, the principle of conservation is not violated when you pass a basketball, nomatter which perspective you take.

4. A prankster puts a firecracker (of negligible mass) into a pumpkin, which then explodes(conveniently) into exactly three pieces. The first piece has a mass of 1.4 kg and initiallymoves West at a speed of 3 m/s. The second piece has a mass of 0.9 kg and initiallymoves North at a speed of 4 m/s. The third piece has a mass of 1.7 kg. Determine themagnitude and direction of the initial velocity of the third piece. [4 points]

Solution: Draw a diagram! Let North and East be the positive directions, as usual.

1Oh, and by the way, would someone please teach the Raptors to move the ball? Some of them seem tohave very sticky hands.

The total initial momentum (i.e., before the explosion) is zero. Thus, the total momen-tum just after the explosion must also be zero. Writing this conservation of momentumprinciple and solving for ~v3, we obtain:

m1~v1 +m2~v2 +m3~v3 = ~0

m3~v3 = −m1~v1 −m2~v2

~v3 =1

m3

[−m1~v1 −m2~v2]

~v3 =1

1.7 kg[− (1.4 kg) (−3 m/s, 0)− (0.9 kg) (0, 4 m/s)]

~v3 =1

1.7 kg[(4.2 kg.m/s, 0)− (0, 3.6 kg.m/s)]

~v3 =1

1.7 kg(4.2,−3.6) kg.m/s

~v3 =1

1.7(4.2,−3.6) m/s

Using Pythagoras’s theorem, the magnitude of the velocity of the third piece of pump-kin is

|~v3| =1

1.7

√4.22 + 3.62 m/s

v3 = 3.3 m/s

The direction of motion is roughly South-East (as you can see from your diagram).Let θ represent the angle between the velocity vector ~v3 and the East direction; then,

tan θ =3.6

4.2

θ = tan−1

(3.6

4.2

)θ = 40.6◦

Thus, the third piece of pumpkin initially moves at a speed of 3.3 m/s in a direction40.6◦ South of East.