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Chapter 7: Problem 36

Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of \(m_{A}=17.0 \mathrm{kg}\) and an initial velocity of \(\overrightarrow{\mathbf{v}}_{a_{A}}=8.00 \mathrm{m} / \mathrm{s},\) due east. Object \(\mathrm{B},\) however, has a mass of \(m_{\mathrm{B}}=29.0 \mathrm{kg}\) and an initial velocity of \(\overrightarrow{\mathbf{v}}_{\mathrm{eg}}=5.00 \mathrm{m} / \mathrm{s},\) due north. Find the magnitude and direction of the total momentum of the two-object system after the collision.

Short Answer

Expert verified
The total momentum magnitude is 198 kg m/s at 46.7 degrees north of east.

Step by step solution

01

Determine Initial Momentum in East Direction

The momentum of object A in the east direction before the collision is given by \( p_{eA} = m_A \cdot v_{iA} = 17.0 \, \text{kg} \times 8.00 \, \text{m/s} = 136.0 \, \text{kg m/s} \). There is no initial momentum contribution from object B in the east direction, so \( p_e = 136.0 \, \text{kg m/s} \).
02

Determine Initial Momentum in North Direction

The momentum of object B in the north direction before the collision is \( p_{nB} = m_B \cdot v_{iB} = 29.0 \, \text{kg} \times 5.00 \, \text{m/s} = 145.0 \, \text{kg m/s} \). Object A contributes nothing in the north direction, so \( p_n = 145.0 \, \text{kg m/s} \).
03

Calculate the Total Momentum Magnitude Using Pythagorean Theorem

The total momentum after the collision is the vector sum of the momenta in the east and north directions. Calculate the magnitude using \( p_{total} = \sqrt{p_e^2 + p_n^2} = \sqrt{(136.0)^2 + (145.0)^2} \, \text{kg m/s} = \sqrt{18496 + 21025} \, \text{kg m/s} \approx 198.0 \, \text{kg m/s} \).
04

Calculate the Direction of Total Momentum

The direction \( \theta \) of the momentum vector is calculated using \( \theta = \arctan\left(\frac{p_n}{p_e}\right) = \arctan\left(\frac{145.0}{136.0}\right) \approx 46.7^\circ \). This angle is measured from the east direction towards the north.

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Key Concepts

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

Momentum Conservation
In a completely inelastic collision, such as the collision between objects A and B, momentum is conserved. This means that despite the objects sticking together after impact, the total momentum of the system before the collision is equal to the total momentum of the system after the collision.

The concept of momentum conservation is critical in solving collision problems. Momentum itself is the product of an object's mass and velocity. In mathematical terms, momentum is expressed as \( p = m \times v \).
  • For object A, the momentum prior to the collision is determined by its mass and velocity in the east direction, \( p_{eA} = m_A \cdot v_{iA} \).
  • For object B, the momentum comes from its movement in the north direction, given by \( p_{nB} = m_B \cdot v_{iB} \).
By calculating these momenta individually, we can ensure that their vector sum remains unchanged after the collision.
Vector Addition
Vector addition is central in calculating the total momentum in collisions involving different directions. In two-dimensional problems, like the collision of objects A and B, you must consider each component of momentum separately.

In our example:
  • Object A adds momentum in the east direction, while object B adds momentum in the north direction, resulting in two perpendicular components of momentum.
  • The total momentum is found through vector addition, which combines these two components using the Pythagorean theorem.
  • The calculation follows: \( p_{total} = \sqrt{p_e^2 + p_n^2} \), where \( p_e \) and \( p_n \) are the momentum components in the east and north directions, respectively.
This method allows us to accurately determine the magnitude of the resulting momentum after collision, taking into account both direction and magnitude.
Direction of Momentum
Understanding the direction of momentum is as important as calculating its magnitude. After a collision, determining the direction of the resultant momentum vector provides insights into the final path of the combined objects.

To find the direction, we use trigonometric relationships:
  • The angle \( \theta \) between the resultant vector and the east direction is calculated with \( \theta = \arctan\left(\frac{p_n}{p_e}\right) \).
  • This formula arises from considering the right triangle formed by the momentum components, where \( p_n \) is the opposite side and \( p_e \) is the adjacent side of the triangle.
In this particular example, \( \theta \approx 46.7^\circ \), indicating that the total momentum vector points northeast, starting from the east axis towards the north. This step is crucial for understanding how the collision changes the system's motion.

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Most popular questions from this chapter

A golf ball strikes a hard, smooth floor at an angle of \(30.0^{\circ}\) and, as the drawing shows, rebounds at the same angle. The mass of the ball is \(0.047 \mathrm{kg},\) and its speed is \(45 \mathrm{m} / \mathrm{s}\) just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball's momentum changes during impact with the floor, and ignore the weight of the ball.)

Two stars in a binary system orbit around their center of mass. The centers of the two stars are \(7.17 \times 10^{11} \mathrm{m}\) apart. The larger of the two stars has a mass of \(3.70 \times 10^{30} \mathrm{kg},\) and its center is \(2.08 \times 10^{11} \mathrm{m}\) from the system's center of mass. What is the mass of the smaller star?

In a science fiction novel two enemies, Bonzo and Ender, are fighting in outer space. From stationary positions they push against each other. Bonzo flies off with a velocity of \(+1.5 \mathrm{m} / \mathrm{s},\) while Ender recoils with a velocity of \(-2.5 \mathrm{m} / \mathrm{s} .\) (a) Without doing any calculations, decide which person has the greater mass. Give your reasoning. (b) Determine the ratio \(m_{\text {Bonno }}\) \(m_{\text {Ender }}\) of the masses of these two enemies.

The lead female character in the movie Diamonds Are Forever is standing at the edge of an offshore oil rig. As she fires a gun, she is driven back over the edge and into the sea. Suppose the mass of a bullet is \(0.010 \mathrm{kg}\) (a) What and its velocity is \(+720 \mathrm{m} / \mathrm{s} .\) Her mass (including the gun) is \(51 \mathrm{kg}\) recoil velocity does she acquire in response to a single shot from a stationary position, assuming that no external force keeps her in place? (b) Under the same assumption, what would be her recoil velocity if, instead, she shoots a blank cartridge that ejects a mass of \(5.0 \times 10^{-4} \mathrm{kg}\) at a velocity of \(+720 \mathrm{m} / \mathrm{s} ?\)

You and your crew must dock your \(25000 \mathrm{kg}\) spaceship at Spaceport Alpha, which is orbiting Mars. In the process, Alpha's control tower has requested that you ram another vessel, a freight ship of mass \(16500 \mathrm{kg},\) latch onto it, and use your combined momentum to bring it into dock. The freight ship is not moving with respect to the colossal Spaceport Alpha, which has a mass of \(1.85 \times 10^{7} \mathrm{kg} .\) Alpha's automated system that guides incoming spacecraft into dock requires that the incoming speed is less than \(2.0 \mathrm{m} / \mathrm{s}\). (a) Assuming a perfectly linear alignment of your ship's velocity vector with the freight ship (which is stationary with respect to Alpha) and Alpha's docking port, what must be your ship's speed (before colliding with the freight ship) so that the combination of the freight ship and your ship arrives at Alpha's docking port with a speed of \(1.50 \mathrm{m} / \mathrm{s} ?\) (b) How does the velocity of Spaceport Alpha change when the combination of your vessel and the freight ship successfully docks with it? (c) Suppose you made a mistake while maneuvering your vessel in an attempt to ram the freight ship and, rather than latching on to it and making a perfectly inelastic collision, you strike it and knock it in the direction of the spaceport with a perfectly elastic collision. What is the speed of the freight ship in that case (assuming your ship had the same initial velocity as that calculated in part (a))?
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