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Chapter 8: Problem 5

One 110 -kg football lineman is running to the right at 2.75 \(\mathrm{m} / \mathrm{s}\) while another 125 -kg lineman is running directly toward him at 2.60 \(\mathrm{m} / \mathrm{s}\) . What are (a) the magnitude and direction of the net momentum of these two athletes, and (b) their total kinetic energy?

Short Answer

Expert verified
The net momentum is \(-22.5\, \text{kg路m/s}\) (left), and the total kinetic energy is \(839.44\, \text{J}\).

Step by step solution

01

Understand the Problem

We have two football linemen running towards each other. We need to find out the net momentum and total kinetic energy of the system. The mass and velocity of each lineman are given. This is a conservation of momentum problem combined with kinetic energy calculation.
02

Define Variables and Equations

Let \(m_1 = 110\, \text{kg}\) and \(v_1 = 2.75\, \text{m/s}\) for the first lineman, and \(m_2 = 125\, \text{kg}\) and \(v_2 = -2.60\, \text{m/s}\) for the second lineman (velocity is negative because he is running towards the first lineman). The net momentum \(p_{net}\) can be calculated as:\[ p_{net} = m_1 \cdot v_1 + m_2 \cdot v_2 \]For kinetic energy, the total kinetic energy \(KE_{total}\) is calculated as:\[ KE_{total} = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 \]
03

Calculate the Net Momentum

Substitute the known values into the equation for net momentum:\[ p_{net} = (110\, \text{kg} \times 2.75\, \text{m/s}) + (125\, \text{kg} \times (-2.60)\, \text{m/s}) \]Solving this, we get:\[ p_{net} = 302.5\, \text{kg路m/s} - 325\, \text{kg路m/s} = -22.5\, \text{kg路m/s} \]Thus, the net momentum is \(-22.5\, \text{kg路m/s}\), directed towards the left (since it is negative).
04

Calculate the Total Kinetic Energy

Substitute the known values into the equation for total kinetic energy:\[ KE_{total} = \frac{1}{2} \times 110\, \text{kg} \times (2.75^2\, \text{m/s}^2) + \frac{1}{2} \times 125\, \text{kg} \times (2.60^2\, \text{m/s}^2) \]Calculating each term:\[ KE_1 = \frac{1}{2} \times 110 \times 7.5625 = 416.9375\, \text{J} \]\[ KE_2 = \frac{1}{2} \times 125 \times 6.76 = 422.5\, \text{J} \]\[ KE_{total} = 416.9375 + 422.5 = 839.4375\, \text{J} \]Hence, the total kinetic energy is \(839.44\, \text{J}\) (rounded to two decimal places).
05

Conclusion and Interpretation

We have found the net momentum and total kinetic energy. The net momentum being negative indicates that the overall momentum is towards the left, meaning the second lineman's impact is stronger. The total kinetic energy provides information about the energy in the system before any possible collision.

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Key Concepts

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

Kinetic Energy Calculation
Kinetic energy is a measure of the energy a body possesses due to its motion. It can be thought of as the energy needed to accelerate an object from rest to a certain velocity. The formula for calculating kinetic energy (\( KE \)) is expressed as:
  • \( KE = \frac{1}{2}mv^2 \)
where \( m \) represents the mass in kilograms and \( v \) is the velocity in meters per second. The kinetic energy is directly proportional to both the mass and the square of the velocity; this means that changes in speed have a more dramatic effect on the kinetic energy compared to changes in mass.
When calculating the total kinetic energy of multiple bodies, such as our football linemen, it's essential to compute the kinetic energy for each individual and then add them up. This gives the total kinetic energy of the system, providing insights into the pre-collision energy dynamics.
Collision Physics
In the realm of physics, collisions are interactions where two or more bodies exert forces on each other in a relatively short time period. A key aspect of collisions is the conservation of momentum. This principle plays a crucial role and states that in the absence of external forces, the total momentum before the collision is equal to the total momentum after the collision.
Given the problem of two football linemen colliding, we assume a closed system where the only significant forces are those exerted by the linemen on each other. Therefore, the momentum they have before they collide is conserved. By calculating this pre-collision momentum, we can predict the outcomes and motion directions post-collision when no external forces act on the system. Understanding these fundamentals clarifies how forces work in sports scenarios and assists in planning smart, safe moves on the field.
Momentum Direction
Momentum in physics refers to the product of a body's mass and its velocity. It is a vector quantity, which means it has both magnitude and direction. The direction of momentum is the same as the direction of the velocity of the moving object.
  • For instance, if an object is moving to the right, its momentum direction is also to the right.
In our example problem, one 110-kg lineman is moving to the right while the 125-kg opponent runs left at differing speeds. This results in a net momentum that can either favor the left or right direction, depending on the relative magnitudes of their individual momenta. After calculations, if the net momentum is negative, it means the predominant force is directed towards the left, indicating stronger influence or momentum from the lineman running left. This directional analysis of momentum gives valuable insight into the effects each player experiences during and after a collision.

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Most popular questions from this chapter

Two identical 1.50-kg masses are pressed against opposite ends of a light spring of force constant \(1.75 \mathrm{N} / \mathrm{cm},\) compressing the spring by 20.0 \(\mathrm{cm}\) from its normal length. Find the speed of each mass when it has moved free of the spring on a frictionless horizontal table.

A blue puck with mass \(0.0400 \mathrm{kg},\) sliding with a velocity of magnitude 0.200 \(\mathrm{m} / \mathrm{s}\) on a frictionless, horizontal air table, makes a perfectly elastic, head-on collision with a red puck with mass \(m,\) initially at rest. After the collision, the velocity of the blue puck is 0.050 \(\mathrm{m} / \mathrm{s}\) in the same direction as its initial velocity. Find (a) the velocity (magnitude and direction) of the red puck after the collision and (b) the mass \(m\) of the red puck.

In a shipping company distribution center, an open cart of mass 50.0 \(\mathrm{kg}\) is rolling to the left at a speed of 5.00 \(\mathrm{m} / \mathrm{s}\) (Fig. P8.95). You can ignore friction between the cart and the floor. A 15.0 -kg package slides down a chute that is inclined at \(37^{\circ}\) from the horizontal and leaves the end of the chute with a speed of 3.00 \(\mathrm{m} / \mathrm{s}\) . The package lands in the cart and they roll off together. If the lower end roll off together. If the lower end of the chute is a vertical distance of 4.00 \(\mathrm{m}\) above the bottom of the cart, what are (a) the speed of the package just before it lands in the cart and (b) the final speed of the cart?

A 10.0 -g marble slides to the left with a velocity of magnitude 0.400 \(\mathrm{m} / \mathrm{s}\) on the frictionless, horizontal surface of an icy New York side- walk and has a head- on, elastic collision with a larger 30.0 -g marble sliding to the right with a velocity of magnitude 0.200 \(\mathrm{m} / \mathrm{s}(\mathrm{Fig} . \mathrm{E} 8.48) .\) (a) Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all the motion is along a line.) (b) Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision for each marble. Compare the values you get for each marble. (c) Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before the collision) for each marble. Compare the values you get for each marble.

A 0.160 -kg hockey puck is moving on an icy, frictionless, horizontal surface. At \(t=0,\) the puck is moving to the right at 3.00 \(\mathrm{m} / \mathrm{s} .\) (a) Calculate the velocity of the puck (magnitude and direction) after a force of 25.0 \(\mathrm{N}\) directed to the right has been applied for 0.050 s. (b) If, instead, a force of 12.0 \(\mathrm{N}\) directed to the left is applied from \(t=0\) to \(t=0.050 \mathrm{s}\) , what is the final velocity of the puck?
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