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Chapter 13: Problem 15

A steel ball of mass \(0.514 \mathrm{~kg}\) is fastened to a cord \(68.7 \mathrm{~cm}\) long and is released when the cord is horizontal. At the bottom of its path, the ball strikes a \(2.63-\mathrm{kg}\) steel block initially at rest on a frictionless surface (Fig. \(13-18\) ). On collision, one-half the mechanical kinetic energy is converted to internal energy and sound energy. Find the final speeds.

Short Answer

Expert verified
The final speeds of the collide steel ball and the initially stationary steel block are obtained using the principles of conservation of energy and conservation of momentum.

Step by step solution

01

Calculate the Initial Velocity of the Ball

Calculate the initial velocity of the ball, when it is at the bottom of its path, just before the collision using the principle of conservation of energy. The potential energy is equal to the kinetic energy which is \(\frac{1}{2} m v^2\), where \(m\) is the mass and \(v\) the velocity. Therefore, \(mgh = \frac{1}{2} m v^2\) where \(g\) is the acceleration due to gravity and \(h\) is the height above the bottom of the path. Solve this equation for \(v\).
02

Use Conservation of Momentum Just Before and After Collision

Just before and after the collision, use the principle of conservation of momentum which states that the total momentum of an isolated system remains constant which is, \(m_1*u_1 = m_1*v_1 + m_2*v_2\), where \(m_1\) and \(m_2\) are the masses of ball and block, \(u_1\) is the initial speed and \(v_1\), \(v_2\) are the final speeds.
03

Use Conservation of Energy After Collision

Use the principle of conservation of energy, just after the collision. According to given problem, half of the kinetic energy of the ball is converted into internal energy and sound. Now compute the final speeds of the ball and block using equations generated from step 2 and step 3.

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Most popular questions from this chapter

A river descends \(15 \mathrm{~m}\) in passing through rapids. The speed of the water is \(3.2 \mathrm{~m} / \mathrm{s}\) upon entering the rapids and is \(13 \mathrm{~m} / \mathrm{s}\) as it leaves. What percentage of the potential energy lost by the water in traversing the rapids appears as kinetic energy of water downstream? What happens to the rest of the energy?

A ball loses \(15.0 \%\) of its kinetic energy when it bounces back from a concrete walk. With what speed must you throw it vertically down from a height of \(12.4 \mathrm{~m}\) to have it bounce back to that same height? Ignore air resistance.

Let the total energy of a system of \(N\) particles be measured in an arbitrary frame of reference, such that \(K=\Sigma \frac{1}{2} m_{n} v_{n}^{2} .\) In the center-of-mass reference frame, the velocities are \(v_{n}^{\prime}=v_{n}-v_{\mathrm{cm}}\), where \(v_{\mathrm{cm}}\) is the velocity of the center of mass relative to the original frame of reference. Keeping in mind that \(v_{n}^{2}=\overrightarrow{\mathbf{v}}_{n} \cdot \overrightarrow{\mathbf{v}}_{n}\), show that the kinetic energy can be written $$K=K_{\mathrm{int}}+K_{\mathrm{cm}}$$ where \(K_{\text {int }}=\Sigma \frac{1}{2} m_{n} v_{n}^{\prime 2}\) and \(K_{\mathrm{cm}}=\frac{1}{2} M v_{\mathrm{cm}}^{2} .\) This demonstrates that the kinetic energy of a system of particles can be divided into an internal term and a center-of-mass term. The internal kinetic energy is measured in a frame of reference in which the center of mass is at rest; for example, the random motions of the molecules of gas in a container at rest are responsible for its internal translational kinetic energy.

While a \(1700-\mathrm{kg}\) automobile is moving at a constant speed of \(15 \mathrm{~m} / \mathrm{s}\), the motor supplies \(16 \mathrm{~kW}\) of power to overcome friction, wind resistance, and so on. (a) What power must the motor supply if the car is to move up an \(8.0 \%\) grade \((8.0 \mathrm{~m}\) vertically for each \(100 \mathrm{~m}\) horizontally) at \(15 \mathrm{~m} / \mathrm{s} ?\) (b) At what downgrade, expressed in percentage terms, would the car coast at \(15 \mathrm{~m} / \mathrm{s}\) ?

An electron, mass \(m\), collides head-on with an atom, mass \(M\), initially at rest. As a result of the collision, a characteristic amount of energy \(E\) is stored internally in the atom. What is the minimum initial speed \(v_{0}\) that the electron must have? (Hint: Conservation principles lead to a quadratic equation for the final electron speed \(v\) and a quadratic equation for the final atom speed \(V\). The minimum value, \(v_{0}\), follows from the requirement that the radical in the solutions for \(v\) and \(V\) be real.)
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