class 9 momentum questions

Check Your Understanding

Express your understanding of the concept and mathematics of momentum by answering the following questions. Click on the button to view the answers.

1. When fighting fires, a firefighter must use great caution to hold a hose that emits large amounts of water at high speeds. Why would such a task be difficult?

 

2. A large truck and a Volkswagen have a head-on collision.

a. Which vehicle experiences the greatest force of impact?

b. Which vehicle experiences the greatest impulse?

c. Which vehicle experiences the greatest momentum change?

d. Which vehicle experiences the greatest acceleration?

 

3. Miles Tugo and Ben Travlun are riding in a bus at highway speed on a nice summer day when an unlucky bug splatters onto the windshield. Miles and Ben begin discussing the physics of the situation. Miles suggests that the momentum change of the bug is much greater than that of the bus. After all, argues Miles, there was no noticeable change in the speed of the bus compared to the obvious change in the speed of the bug. Ben disagrees entirely, arguing that that both bug and bus encounter the same force, momentum change, and impulse. Who do you agree with? Support your answer.

 

4. If a ball is projected upward from the ground with ten units of momentum, what is the momentum of recoil of the Earth? ____________ Do we feel this? Explain.

 

5. If a 5-kg bowling ball is projected upward with a velocity of 2.0 m/s, then what is the recoil velocity of the Earth (mass = 6.0 x 10²⁴ kg).

 

6. A 120 kg lineman moving west at 2 m/s tackles an 80 kg football fullback moving east at 8 m/s. After the collision, both players move east at 2 m/s. Draw a vector diagram in which the before- and after-collision momenta of each player is represented by a momentum vector. Label the magnitude of each momentum vector.

See answer below.

7. In an effort to exact the most severe capital punishment upon a rather unpopular prisoner, the execution team at the Dark Ages Penitentiary search for a bullet that is ten times as massive as the rifle itself. What type of individual would want to fire a rifle that holds a bullet that is ten times more massive than the rifle? Explain.

 

8. A baseball player holds a bat loosely and bunts a ball. Express your understanding of momentum conservation by filling in the tables below.

9. A Tomahawk cruise missile is launched from the barrel of a mobile missile launcher. Neglect friction. Express your understanding of momentum conservation by filling in the tables below.

Answer to Question #6

Return to question #6.

Next Section: Isolated Systems
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maths pythagoras proofs

The Pythagorean Theorem states:

In a right triangle, with sides (legs) a and b, and hypotenuse c, then c²=a²+b².

A right triangle is a triangle with one right angle (an angle of 90°). Its hypotenuse is the side opposite the right angle.
Proof #1: The simplest proof is an algebraic proof using similar triangles ABC, CBX, and ACX (in the diagram):

Since corresponding parts of similar triangles are proportional, a/x=c/a or a²=cx. And b/(c-x)=c/b or b²=c²-cx or c²=cx+b². Substituting a² for cx, we get c²=a²+b². Which is what we were trying to prove.

This proof is by Legendre, and was probably originally devised by an ancient Hindu mathematician. Euclid's proof is quite a bit more complicated than that. It is actually surprising that he did not come up with a proof similar to the above. But, his proof is clever, as well.

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Proof #2: Here is another nice proof:
We start with a right triangle (in gold, in the diagram) with sides a, b, and c. We then build a big square, out of four copies of our triangle, as shown at the left. We end up with a square, in the middle, with sides c (we can easily show that this is a square).
We now construct a second big square, with identical triangles which are arranged as in the lower part of the diagram. This square has the same area as the square above it.
We now sum up the parts of the two big squares:

Area=2ab + c²
Area=2ab + a² + b²

These two areas are equal:

2ab + c²=2ab + a² + b²
c²=a² + b²

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Proof #3: This diagram might look familiar. I've just drawn the squares on the sides of our right triangle. And, I've drawn a line from the right angle of the triangle, perpendicular to the hypotenuse, through the square which is on the hypotenuse. The idea is to prove that the little square (in blue) has the same area as the little rectangle (also in blue). I've named the width of this rectangle, x.
Using similar triangles, within our right triangle:

a/c=x/a
x=a²/c

The area of the blue rectangle is xc. So we just plug in a²/c for x:

area=(a²/c)c=a²

So, the area of the blue rectangle is equal to the area of the blue square. Similarly, the area of the other rectangle (to the left of the blue one) is equal in area of the other square (b²).
So, the total area of the big square is:

c²=a² + b²

That was actually fairly easy. I made that proof up, as I studied Euclid's proof, below.

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Proof #4: We now look at Euclid's proof. Although he proved many of the theorems of algebra in his books, he did not have the methods of algebra (the manipulation of symbols) available to him. So, we must add a few more lines. We label a few points (see the diagram), and draw lines BI and CE.
Triangle ACE is congruent to triangle AIB, by the side-angle-side theorem (it's a postulate, nowadays), because AC=AI, AE=AB and angle CAE=angle IAB (both angles are congruent to angle CAB + a right angle). [Note: I am using the=sign for congruence. And I am labelling a line segment as AC, because I lack more appropriate symbols].
But, the area of triangle AIB is equal to half the area of square ACHI (They have the same base AI and height IH), and the area of triangle ACE is equal to half the area of the rectangle AGFE. So, the square ACHI is equal, in area, to the rectangle AGFE. Likewise, we show that the area of the smaller square is equal to the area of the smaller rectangle.
So, the area of the large square is equal to the sums of the areas of the two smaller squares. So, we've proved the Pythagorean theorem, again.

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Here is a square with two diagonals. The area of the four triangles is 2a². The area of the square is c². Obviously a²+a²=c², a special case of the Pythagorean Theorem. If a is 1, then c is sqrt(2) [the square root of 2]. Or if c is 1, then a is sqrt(2)/2. This diagram does not seem to help in proving the Pythagorean Theorem.

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Mathematics historians are fairly certain that the Pythagorean Theorem was known before Pythagoras. Various cultures seem to have known the length of the diagonal of a square (sqrt(2)) fairly accurately, more accurately than they could measure with a ruler. See The 3-4-5 Right Triangle In Ancient Egypt, where I express the opinion that knowledge of a 3-4-5 right triangle does not imply knowledge of the Pythagorean Theorem.
The book The Pythagorean Proposition, By Elisha Scott Loomis, is a fairly amazing book. It contains 256 proofs of the Pythagorean Theorem. It shows that you can devise an infinite number of algebraic proofs, like the first proof above. It shows that you can devise an infinite number of geometric proofs, like Euclid's proof. And it shows that there can be no proof using trigonometry, analytic geometry, or calculus. The book is out of print, by the way.

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The Pythagorean Theorem depends on the parallel postulate. So, it does not hold for the non-Euclidean geometries. In relation to the above proofs

1. There are no similar triangles in the non-Euclidean geometries.
2. We cannot prove that the above triangles have half the area of the square or rectangle.

We also have the problem that there are no squares in the non-Euclidean geometries.

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Note: Congruence applies to geometric figures, including line segments and angles. Equality applies to numbers, such as areas, lengths of line segments, and the measures of angles. A square and a rectangle are not equal, their areas are equal.
See Rik Littlefield's proof, which is interesting. Also, see The 3-4-5 Right Triangle In Ancient Egypt. The Pythagorean Theorem is also mentioned in Isosceles Triangles.

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Addendum #1:
Proof #5: This proof is attributed to President Garfield. Given the right triangle at the bottom of this diagram, duplicate it as shown, and complete the trapezoid as shown. The larger triangle (with two sides c) is a right triangle (fairly easy to show). The areas of the three triangles are ab/2, ab/2 and c²/2. The sum of these is ab+c²/2. The area of a trapezoid is 1/2 the sum of the parallel sides times the height. Here that is (a+b)(a+b)/2 or (a²+2ab+b²)/2. These two expressions (for the three triangles and the trapezoid) represent the same area:

ab+c²/2=(a²+2ab+b²)/2
2ab+c²=a²+2ab+b² [double both sides]
c²=a²+b²

This proof has a great deal in common with our second proof, above.

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Proof #6: This proof is attributed to Bhaskara, a Hindu mathematician of the 12th century. We are given the bottom right triangle. We construct a square by making three copies of the triangle, as shown. We assume that a is the smallest side, less than or equal to b. The area of the large square is c². The side of the small square is b-a, and its area is (b-a)² or b²-2ab+a². The area of our triangle is ab/2. The area of all four triangles is 2ab. Then the area of all four triangles, plus the area of the small square is b²+a². So c²=a²+b².

force of friction

Friction is a force that is created whenever two surfaces move or try to move across each other. 

• Friction always opposes the motion or attempted motion of one surface across another surface.
• Friction is dependant on the texture of both surfaces.
• Friction is also dependant on the amount of contact force pushing the two surfaces together (normal force). 

In this simulation you see a block sitting on a level table.  You can place an applied force on the object by pressing the "More Force" button.   Each time you press the button the applied force will increase.  As you use this simulation there are several things you should notice.

• On a level surface, the normal force (F_N) is always equal and opposite to the weight (only on a level surface).
• The force of static friction (f_s) cancels out the applied force right up to and including when static friction reaches its maximum (f_smax).
• For applied forces greater than the maximum force of static friction the block starts to slip and then the value for friction becomes kinetic friction (f_k) and the box is then under a net force so it accelerates to the right.

The force of friction depends upon both surfaces in contact and the normal force.  A mathematical relationship can be created here.

	In this first example, a block of wood is shown sliding across a wooden table.  (notice the cause of this sliding is not shown)  Notice that the force of kinetic friction (fk) is equal to 40% of the normal force (FN).   Another way of writing this relationship would be
	Now as we compare the first simulation to the next, we find that if the weight of the block is doubled, the normal force doubles, and the force of friction becomes doubled.  Once again we find that the force of kinetic friction (fk) is equal to 40% of the normal force (FN).  Another way of writing this relationship would be    Since this value is true for any weight of wood on wood, we say this value represents the coefficient of friction.  (It's just the percentage of the normal force that can be friction.)The formula for the coefficient of kinetic friction is
	As we compare the simulation of wood on wood to wood on asphalt, we find that the amount of friction on the block increased for for the same amount of weight.  Notice that the force of kinetic friction (fk) is equal to 60% of the normal force (FN) or we could say
	Another coefficient can be used to describe the relationship between the maximum force of static friction and the normal force.  It's called the coefficient of static friction and its formula looks like    The maximum force of static friction is used because static friction has a whole range from zero newtons up to the maximum force of static friction.

The coefficients for static and kinetic friction are listed in some reference tables.  The coefficient of static friction is usually a little bit higher than coefficient of kinetic friction for the same two surfaces.  When coefficients are listed they must be given for one surface on another surface (ie wood-on-asphalt).  The higher the coefficient, the greater the force of friction.   The table below lists the coefficients for a few common surfaces used in physics.   They are arranged from "sticky" to slippery.

surface-on-surface
hook velcro-on-fuzzy velcro	>6.0	>5.9
avg tire-on-dry pavement	0.9	0.8
grooved tire-on-wet pavement	0.8	0.7
glass-on-glass	0.9	0.4
metal-on-metal (dry)	0.6	0.4
smooth tire-on-wet pavement	0.5	0.4
metal-on-metal (lubricated)	0.1	0.05
steel-on-ice	0.1	0.05
steel-on-Teflon	0.05	0.05
You should keep in mind that it isn't possible to give accurate values for the coefficient of frictions due to changing surface smoothness.  For example, not all pieces of metal have the same surface smoothness.  Some that are highly polished may be more slippery than others that are pitted or scratched.  These values are just meant to give you the approximate values.

PHYSICS

IF force = mass.accelaration 
then force = mass.k.accelaration
,where k is constant
then why k is not 1 in each case?!?