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Two skaters collide and grab on to each other on frictionless ice. One of them, of mass 70.0 kg, is moving to the right at 4.00 m/s, while the other, of mass 65.0 kg, is moving to the left at 2.50 m/s. What are the magnitude and direction of the velocity of these skaters just after they collide?

Short Answer

Expert verified
The skaters move together at 0.87 m/s to the right after the collision.

Step by step solution

01

Understanding the Conservation of Momentum

In a collision, where no external forces are acting on the system, the total momentum before the collision is equal to the total momentum after the collision, according to the law of conservation of momentum. We need to calculate the total momentum before the collision to find the velocity of the combined masses after they collide.
02

Calculating Initial Momentum of Each Skater

The momentum of an object is given by the product of its mass and velocity. For the first skater (70.0 kg moving at 4.00 m/s to the right):\[ p_1 = 70.0 \text{ kg} \times 4.00 \text{ m/s} = 280.0 \text{ kg m/s to the right}\]For the second skater (65.0 kg moving at 2.50 m/s to the left):\[ p_2 = 65.0 \text{ kg} \times (-2.50 \text{ m/s}) = -162.5 \text{ kg m/s to the left}\](Note: Velocity to the left is negative.)
03

Calculating Total Initial Momentum

The total initial momentum of the system is the sum of the momenta of the two skaters:\[ p_{\text{total, initial}} = p_1 + p_2 = 280.0 \text{ kg m/s} + (-162.5 \text{ kg m/s}) = 117.5 \text{ kg m/s}\]
04

Calculating the Total Mass After Collision

After the collision, the skaters grab onto each other and move together as a single object. Therefore, the total mass of the system after the collision is the sum of their masses:\[ m_{\text{total}} = 70.0 \text{ kg} + 65.0 \text{ kg} = 135.0 \text{ kg}\]
05

Finding the Velocity After Collision

Using the conservation of momentum, where total momentum before collision equals total momentum after collision:\[ p_{\text{total, initial}} = m_{\text{total}} \times v_{\text{final}}\]Substitute the known values:\[ 117.5 \text{ kg m/s} = 135.0 \text{ kg} \times v_{\text{final}}\]Solve for \( v_{\text{final}} \):\[ v_{\text{final}} = \frac{117.5 \text{ kg m/s}}{135.0 \text{ kg}} = 0.87 \text{ m/s}\]
06

Analyzing Direction of Motion

Since the total initial momentum is positive, the final direction of motion is to the right. The final velocity of the pair moving together is 0.87 m/s to the right.

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Key Concepts

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

Inelastic Collision
In physics, an inelastic collision is a type of collision where the colliding objects stick together after the impact. Unlike an elastic collision, where the objects bounce apart and total kinetic energy is conserved, inelastic collisions do not conserve kinetic energy. Instead, they conserve momentum.
  • Inelastic collisions often result in deformations or energy being lost as heat or sound.
  • The focus is on the conservation of the total momentum of the system.
In the context of the skaters on ice, after colliding, they hold onto each other and travel together as a single body. This behavior is a signature of a perfectly inelastic collision, demonstrating that even though energy is not conserved, we can predict outcomes using momentum conservation principles.
Momentum
Momentum is a fundamental concept in physics describing the quantity of motion an object has. It is calculated as the product of an object's mass and its velocity. Expressed mathematically, momentum $p = m imes v$, where $m$ is mass and $v$ is velocity.
  • It is a vector quantity, meaning it has both magnitude and direction.
  • The law of conservation of momentum states that in a closed system with no external forces, the total momentum remains constant.
For the skaters problem, each skater has its momentum calculated individually before the collision. The total system momentum is then calculated by adding these vectors, taking care of their respective directions (right is positive, left is negative). This total allows predicting the system's behavior after the collision.
Physics Problems
Solving physics problems often requires a systematic approach. Understanding the foundational principles such as momentum conservation is crucial. Let's break down the key steps you can take:
  • Clearly identify what is given and what needs to be found.
  • Use appropriate physical laws. For example, here we use the conservation of momentum because there are no external forces acting on the skaters.
  • Translate the problem into mathematical equations, ensuring correct sign conventions for directions.
  • Solve equations carefully using algebraic manipulation.
Applying these principles to the skaters, we step through calculating each skater's momentum, summing them to find total momentum, and using this to find their combined velocity after collision. This structured approach is valuable for solving a wide range of physics problems beyond collisions.

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Most popular questions from this chapter

A machine part consists of a thin, uniform 4.00-kg bar that is 1.50 m long, hinged perpendicular to a similar vertical bar of mass 3.00 kg and length 1.80 m. The longer bar has a small but dense 2.00-kg ball at one end (\(\textbf{Fig. E8.55}\)). By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivoted counterclockwise through 90\(^\circ\) to make the entire part horizontal?

To keep the calculations fairly simple but still reasonable, we model a human leg that is 92.0 cm long (measured from the hip joint) by assuming that the upper leg and the lower leg (which includes the foot) have equal lengths and are uniform. For a 70.0-kg person, the mass of the upper leg is 8.60 kg, while that of the lower leg (including the foot) is 5.25 kg. Find the location of the center of mass of this leg, relative to the hip joint, if it is (a) stretched out horizontally and (b) bent at the knee to form a right angle with the upper leg remaining horizontal.

A 15.0-kg fish swimming at 1.10 m/s suddenly gobbles up a 4.50-kg fish that is initially stationary. Ignore any drag effects of the water. (a) Find the speed of the large fish just after it eats the small one. (b) How much mechanical energy was dissipated during this meal?

A 12.0-kg shell is launched at an angle of 55.0\(^\circ\) above the horizontal with an initial speed of 150 m/s. At its highest point, the shell explodes into two fragments, one three times heavier than the other. The two fragments reach the ground at the same time. Ignore air resistance. If the heavier fragment lands back at the point from which the shell was launched, where will the lighter fragment land, and how much energy was released in the explosion?

Blocks \(A\) (mass 2.00 kg) and \(B\) (mass 10.00 kg, to the right of \(A\)) move on a frictionless, horizontal surface. Initially, block \(B\) is moving to the left at 0.500 m/s and block \(A\) is moving to the right at 2.00 m/s. The blocks are equipped with ideal spring bumpers, as in Example 8.10 (Section 8.4). The collision is headon, so all motion before and after it is along a straight line. Find (a) the maximum energy stored in the spring bumpers and the velocity of each block at that time; (b) the velocity of each block after they have moved apart.

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