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Chapter 9: Problem 40

One billiard ball is shot east at \(2.0 \mathrm{m} / \mathrm{s} .\) A second, identical billiard ball is shot west at \(1.0 \mathrm{m} / \mathrm{s}\). The balls have a glancing collision, not a head-on collision, deflecting the second ball by \(90^{\circ}\) and sending it north at \(1.41 \mathrm{m} / \mathrm{s} .\) What are the speed and direction of the first ball after the collision? Give the direction as an angle south of cast.

Short Answer

Expert verified
The speed of the first ball after the collision is \(3.36 m/s\) and the direction is \(25.4\) degrees south of east.

Step by step solution

01

Identify the initial momenta

The first ball P1 is moving east at 2.0 m/s and the second ball P2 is moving west at 1.0 m/s. Therefore, we have the initial vectors as: \( P1_{init} = 2.0i \), \( P2_{init} = -1.0i \). The use of \(i\) and \(j\) refer to the east and north directions respectively.
02

Identify the final momenta

After the collision, the second ball is moving north at 1.41 m/s. The first ball is deflected to an unknown direction. We denote the final momentum of first ball as \(P1_{final}\). The final vectors are: \( P1_{final} = xi + yj \) (the momentum of the first ball which we need to find) and \( P2_{final} = 1.41j \). Where x and y will be components of the final speed of the first ball.
03

Applying the principle of conservation of momentum in horizontal and vertical directions

According to the principle of conservation of momentum, the sum of the initial momenta should be equal to the the sum of the final momenta. Hence we can write two equations for x and y directions. For x direction: \(2.0 = x - 1.0\), hence, \(x = 3.0 m/s\). For y direction: \(0 = y + 1.41\), hence \(y = -1.41 m/s\). The negative sign indicated the direction is towards south.
04

Calculate the final speed and direction of the first ball

The final speed (magnitude of momentum) of the first ball could be found using the Pythagorean rule: \(|P1_{final}| = \sqrt{x^2 + y^2} = \sqrt{(3.0)^2 + (-1.41)^2} = 3.36 m/s\). The direction 螛 can be found using arctangent of the y-component divided by the x-component: 螛 = arctan(|y|/x) = arctan(1.41/3.0) = 25.4 degrees. The direction is south of east.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Collision
When two objects collide, they interact with each other through a force that affects their motion. In our example, two billiard balls meet in a glancing collision instead of a direct head-on collision.
A glancing collision involves the objects coming together at an angle that isn't parallel, causing a change in direction or spin rather than merely a change in speed.
  • This type of collision often results in one or both objects being deflected in new directions.

  • Here, the second ball, originally moving west, is deflected north at 90 degrees.

  • Understanding the type of collision helps in predicting the resulting paths and speeds of the objects involved.

Using the conservation of momentum, we can further analyze the effects of such a collision on both objects.
Momentum
Momentum is a core concept in physics described by the equation: \( p = mv \), where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity of an object.
Momentum is conserved in collisions, meaning the total momentum before the collision equals the total momentum after.
  • In this example, both billiard balls are identified with initial momenta based on their velocities.

  • The conservation law allows us to write equations that equate the sum of initial and final momenta vectors for each direction.

  • By solving these equations, we determine the unknown momentum components of the first ball after the collision.

These principles of momentum conservation, both horizontally and vertically, provide the framework to solve broader questions in mechanics.
Vectors
Vectors are mathematical objects used to represent quantities with both magnitude and direction, such as velocity and momentum in this problem.
The eastward and northward directions define the vector components, typically represented as \( i \) and \( j \), respectively.
  • The first ball's eastward motion is initially simple, represented by a vector in the positive \( i \) direction.

  • When the collision causes changes, the final momentum vectors may include both \( i \) and \( j \) components, signifying different speeds and directions.

  • These components help visualize and solve for the ball's new velocity and path.

Using vectors provides a comprehensive way to analyze and compute the effects of forces and movements in two dimensions.
Pythagorean Theorem
The Pythagorean Theorem is crucial for finding the magnitude of vectors in this collision scenario, which results in orthogonal vector components.
  • Once the components \( x \) and \( y \) of the final velocity are determined, the theorem helps calculate the resultant speed.

  • This theorem states: if a right triangle has sides \( a \), \( b \), and hypotenuse \( c \), then \( c = \sqrt{a^2 + b^2} \).

  • For the first ball's post-collision speed, we calculate \( \|P1_{final}\| = \sqrt{(3.0)^2 + (-1.41)^2} = 3.36 \text{ m/s} \).

Additionally, using trigonometric functions like arctangent helps find the direction of a resultant vector, enhancing the understanding of the paths followed after collisions.

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Most popular questions from this chapter

You are a world-famous physicist-lawyer defending a client who has been charged with murder. It is alleged that your client, Mr. Smith, shot the victim, Mr. Wesson. The detective who investigated the scene of the crime found a second bullet, from a shot that missed Mr. Wesson, that had embedded itself into a chair. You arise to cross-examine the detective. You: In what type of chair did you find the bullet? Det: A wooden chair. You: How massive was this chair? Det: It had a mass of \(20 \mathrm{kg}\).You: How did the chair respond to being struck with a bullet? Det: It slid across the floor. You: How far? Det: Three centimeters. The slide marks on the dusty floor are quite distinct. You: What kind of floor was it? Det: A wood floor, very nice oak planks. You: What was the mass of the bullet you retrieved from the chair? Det: Its mass was \(10 \mathrm{g}\). You: And how far had it penetrated into the chair? Det: A distance of \(4 \mathrm{cm}\) You: Have you tested the gun you found in Mr. Smith's possession? Det: I have. You: What is the muzzle velocity of bullets fired from that gun? Det: The muzzle velocity is \(450 \mathrm{m} / \mathrm{s}\) You: And the barrel length? Det: The gun has a barrel length of \(62 \mathrm{cm} .\) With only a slight hesitation, you turn confidently to the jury and proclaim, "My client's gun did not fire these shots!" How are you going to convince the jury and the judge?

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A 50kg archer, standing on frictionless ice, shoots a \(100 \mathrm{g}\) arrow at a speed of \(100 \mathrm{m} / \mathrm{s}\). What is the recoil speed of the archer?

Two 500 g blocks of wood are \(2.0 \mathrm{~m}\) apart on a frictionless table. A \(10 \mathrm{~g}\) bullet is fired at \(400 \mathrm{~m} / \mathrm{s}\) toward the blocks. It passes all the way through the first block, then embeds itself in the second block. The speed of the first block immediately afterward is \(6.0 \mathrm{~m} / \mathrm{s}\). What is the speed of the secand block after the bullet stops in it?

A neutron is an electrically neutral subatomic particle with a mass just slightly greater than that of a proton. A free neutron is radioactive and decays after a few minutes into other subatomic particles. In one experiment, a neutron at rest was observed to decay into a proton (mass \(1.67 \times 10^{-27} \mathrm{kg}\) ) and an electron (mass \(9.11 \times 10^{-31} \mathrm{kg}\) ). The proton and electron were shot out back-to-back. The proton speed was measured to be \(1.0 \times\) \(10^{5} \mathrm{m} / \mathrm{s}\) and the electron speed was \(3.0 \times 10^{7} \mathrm{m} / \mathrm{s} .\) No other decay products were detected. a. Was momentum conserved in the decay of this neutron? NOTE \(\rightarrow\) Experiments such as this were first performed in the \(1930 \mathrm{s}\) and seemed to indicate a failure of the law of conservation of momentum. In \(1933,\) Wolfgang Pauli postulated that the neutron might have a thind decay product that is virtually impossible to detect. Bven so, it can carry away just enough momentum to keep the total momentum conserved. This proposed particle was named the neutrino, meaning "little neutral one." Neutrinos were, indeed, discovered nearly 20 years later. b. If a neutrino was emitted in the above neutron decay, in which direction did it travel? Explain your reasoning. c. How much momentum did this neutrino "carry away" with it?
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