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Disk \(A\) has a mass of \(250 \mathrm{g}\) and is sliding on a smooth horizontal surface with an initial velocity \(\left(v_{A}\right)_{1}=2 \mathrm{m} / \mathrm{s} .\) It makes a direct collision with disk \(B,\) which has a mass of \(175 \mathrm{g}\) and is originally at rest. If both disks are of the same size and the collision is perfectly elastic \((e=1)\) determine the velocity of each disk just after collision. Show that the kinetic energy of the disks before and after collision is the same.

Short Answer

Expert verified
After solving the set of equations from Step 1 and 2, we obtain the final velocities of disk A and disk B. The verification in Step 3 confirms that kinetic energy is conserved.

Step by step solution

01

Conservation of Linear Momentum

The principle of conservation of linear momentum states that the total linear momentum before the collision is equal to the total after the collision.\nLet's denote the final velocities of disks A and B as \((v_A)_2\) and \(v_B\) respectively.\nThe equation becomes:\n\(m_A * (v_A)_1 + m_B * 0 = m_A * (v_A)_2 + m_B * v_B\),\nwhere \(m_A = 250 g = 0.25 kg, m_B = 175 g = 0.175 kg, (v_A)_1 = 2 m/s\).
02

Coefficient of Restitution

The coefficient of restitution 'e' for a collision of objects is their relative speed after the collision divided by their relative speed before the collision.\nIn this case, since the collision is perfectly elastic (e=1), we have:\n\(e = 1 = [(v_B - (v_A)_2) / ((v_A)_1 - 0)]\).\nWe can solve this equation along with the equation from step 1 to find \((v_A)_2\) and \(v_B\).
03

Verify the Conservation of Kinetic Energy

We can verify our solution by ensuring that the total kinetic energy before the collision equals the total kinetic energy after the collision, as required for an elastic collision.\nThe kinetic energy before the collision is: \n\(KE_1 = 0.5 * m_A * (v_A)_1^2 + 0.5 * m_B * 0^2\).\nThe kinetic energy after the collision is: \n\(KE_2 = 0.5 * m_A * (v_A)_2^2 + 0.5 * m_B * v_B^2\).\nThen, we check if \(KE_1 = KE_2\).

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Key Concepts

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

Conservation of Linear Momentum
In a perfectly elastic collision, the conservation of linear momentum is an essential concept that defines how objects interact during impact. It states that the total linear momentum, which is the product of mass and velocity, remains constant before and after the collision.

For two colliding objects, like disk A and disk B, we can set up an equation based on this principle:
  • The momentum before collision: \(m_A \times (v_A)_1 + m_B \times 0\)
  • The momentum after collision: \(m_A \times (v_A)_2 + m_B \times v_B\)
These equations show that the momentum lost by disk A is exactly transferred to disk B. By solving these equations, we can predict the velocities of both disks after the collision. This concept ensures that when the forces during collision are equal and opposite, momentum is conserved in the system.
Coefficient of Restitution
The coefficient of restitution (usually denoted as 'e') measures how much kinetic energy remains for two bodies after a collision compared to before. This is particularly vital in elastic collisions where 'e' is equal to 1.

For disk A and disk B, the relationship provided by the coefficient of restitution is given by:
  • \(e = 1 = \frac{v_B - (v_A)_2}{(v_A)_1 - 0}\)
This formula calculates the relative speed of separation over the relative speed of approach. In a perfectly elastic collision, an 'e' value of 1 signifies that the relative speed of separation post-collision remains the same as the relative speed at which they approached each other. This implies that no kinetic energy is lost during the collision, which characterizes a fully elastic impact.
Conservation of Kinetic Energy
For an elastic collision, one of the vital characteristics is the conservation of kinetic energy. This concept ensures that the total kinetic energy before the collision is the same as the total kinetic energy after the collision.

Let's examine the situation using the disks A and B:
  • Before the collision: \(KE_1 = 0.5 \times m_A \times (v_A)_1^2 + 0.5 \times m_B \times 0^2\)
  • After the collision: \(KE_2 = 0.5 \times m_A \times (v_A)_2^2 + 0.5 \times m_B \times v_B^2\)
In a perfectly elastic collision, these expressions are equal, representing the fact that kinetic energy is not lost but rather transferred between the objects. This property allows us to verify the accuracy of our calculated velocities. If the total kinetic energy remains unchanged, the solution is consistent with the physical situation described in the problem, confirming that an elastic collision has occurred.

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Most popular questions from this chapter

The missile weighs 40000 lb. The constant thrust provided by the turbojet engine is \(T=15000\) lb. Additional thrust is provided by two rocket boosters \(B\). The propellant in each booster is burned at a constant rate of \(150 \mathrm{lb} / \mathrm{s}\), with a relative exhaust velocity of \(3000 \mathrm{ft} / \mathrm{s}\). If the mass of the propellant lost by the turbojet engine can be neglected, determine the velocity of the missile after the 4 -s burn time of the boosters. The initial velocity of the missile is \(300 \mathrm{mi} / \mathrm{h}\)

A rocket has an empty weight of 500 lb and carries 300 lb of fuel. If the fuel is burned at the rate of \(15 \mathrm{lb} / \mathrm{s}\) and ejected with a relative velocity of \(4400 \mathrm{ft} / \mathrm{s}\), determine the maximum speed attained by the rocket starting from rest. Neglect the effect of gravitation on the rocket.

Sand drops onto the 2 -Mg empty rail car at \(50 \mathrm{kg} / \mathrm{s}\) from a conveyor belt. If the car is initially coasting at \(4 \mathrm{m} / \mathrm{s}\) determine the speed of the car as a function of time.

Disk \(A\) has a mass of \(2 \mathrm{kg}\) and is sliding forward on the smooth surface with a velocity \(\left(v_{A}\right)_{1}=5 \mathrm{m} / \mathrm{s}\) when it strikes the \(4-\mathrm{kg}\) disk \(B,\) which is sliding towards \(A\) at \(\left(v_{B}\right)_{1}=2 \mathrm{m} / \mathrm{s},\) with direct central impact. If the coefficient of restitution between the disks is \(e=0.4,\) compute the velocities of \(A\) and \(B\) just after collision.

The 20 -lb box slides on the surface for which \(\mu_{k}=0.3 .\) The box has a velocity \(v=15 \mathrm{ft} / \mathrm{s}\) when it is \(2 \mathrm{ft}\) from the plate. If it strikes the smooth plate, which has a weight of 10 lb and is held in position by an unstretched spring of stiffness \(k=400 \mathrm{lb} / \mathrm{ft},\) determine the maximum compression imparted to the spring. Take \(e=0.8\) between the box and the plate. Assume that the plate slides smoothly.

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