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Chapter 8: Problem 5

One \(110 \mathrm{~kg}\) football lineman is running to the right at \(2.75 \mathrm{~m} / \mathrm{s}\) while another 125 kg lineman is nunning directly toward him at \(2,60 \mathrm{~m} / \mathrm{s}\). What ure (a) the magnitude and direction of the net momentum of these two athletes, and (b) their total kinctic encrgy?

Short Answer

Expert verified
The total momentum of the football players is calculated to be the value of \(p\) from 'Step 2' and in the direction of the player with positive momentum. The total kinetic energy of the athletes will be the value of 'KE' from 'Step 3'

Step by step solution

01

Identify the parameters

We've been given the mass and velocity of two football players. Let's denote the first player with subscript 1 (mass \(m_1 = 110kg\) and velocity \(v_1 = 2.75 m/s\) to the right) and the second player with subscript 2 (mass \(m_2 = 125kg\) and velocity \(v_2 = 2.60 m/s\) to the left). Since opposite directions are considered, it is important to assign opposite signs to the velocities.
02

Calculate the net momentum

The total momentum (\(p\)) of a system is the vector sum of the individual momenta. For one dimension case like this, it can be obtained by summing the product of mass and velocity of each player. So, \(p = m_1 * v_1 + m_2 * v_2\). Substitute the given values and solve for \(p\).
03

Calculate the total kinetic energy

The kinetic energy (KE) is given by the formula \(KE = 1/2 * m * v^2\). The total kinetic energy is the sum of the kinetic energy of each player. Therefore, \(KE = 1/2 * m_1 * v_1^2 + 1/2 * m_2 * v_2^2\). Substitute the given values and solve for KE.

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Key Concepts

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

Kinetic Energy
Kinetic energy (KE) is the energy an object possesses due to its motion. It can be quantified using the formula ewlineewlineewlineewline
Conservation of Momentum
The law of conservation of momentum states that in a closed system, without external forces, the total momentum remains constant. It's crucial in solving problems involving collisions or explosions. If we observe two football players running towards each other and they collide, unless acted on by an external force, their total momentum just before the collision will equal the total momentum just after.

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Most popular questions from this chapter

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In Section 8.5 we calculated the center of mass by considering objects composed of a finite number of point masses or objects that, by symmetry, could be represented by a finite number of point masses. For a solid object whose mass distribution does not allow for a simple determination of the center of mass by symmetry, the sums of Eqs. (8.28) must be generalized to integrals $$ x_{\mathrm{cm}}=\frac{1}{M} \int x d m \quad y_{\mathrm{cm}}=\frac{1}{M} \int y d m $$ where \(x\) and \(y\) are the coordinates of the small picce of the object that has mass \(d m\). The integration is over the whole of the object. Consider a thin rod of length \(I_{4}\), mass \(M\), and cross-sectional area \(A\). I et the origin of the coordinates be at the left end of the rod and the positive \(x\) -axis lie along the rod. (a) If the density \(\rho=M / V\) of the object is uniform, perform the integration described above to show that the \(x\) -coordinate of the center of mass of the rod is at its gcometrical center. (b) If the density of the object varies linearly with \(x\) - that is, \(\rho=\alpha x,\) where \(\alpha\) is a positive constant - calculate the \(x\) -coordinate of the rod's center of mass.

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