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Chapter 8: Problem 12

A pendulum consisting of a ball of mass \(m\) is released from the position shown in Fig. \(8-2\) and strikes a block of mass \(M\). The block slides a distance \(D\) before stopping under the action of a steady friction force of \(0.20 M g .\) Find \(D\) if the ball rebounds to an angle of \(20^{\circ}\). The pendulum ball falls through a height \(\left(L-L \cos 37^{\circ}\right)=0.201 L\) and rebounds to a height \(\left(L-L \cos 20^{\circ}\right)=\) \(0.0603 L\). Because \((m g h)_{\text {top }}=\left(\frac{1}{2} m v^{2}\right)_{\text {bottom }}\) for the ball, its speed at the bottom is \(v=\sqrt{2 g h}\). Thus, just before it hits the block, the ball has a speed equal to \(\sqrt{2 g(0.201 L)}\). Since the ball rises up to a height of \(0.0603 L\) after the collision, it must have rebounded with an initial speed of \(\sqrt{2 g(0.0603 L)}\). Although \(\mathrm{KE}\) is not conserved in the collision, momentum is. Therefore, for the collision, Momentum just before \(=\) Momentum just after $$ m \sqrt{2 g(0.201 L)}+0=-m \sqrt{2 g(0.0603 L)}+M V $$ where \(V\) is the velocity of the block just after the collision. (Notice the minus sign on the momentum of the rebounding ball.) Solving this equation, we find $$ V=\frac{m}{M} 0.981 \sqrt{g L} $$ The block uses up its translational KE doing work against friction as it slides a distance \(D\). Therefore, $$ \frac{1}{2} M V^{2}=F_{\mathrm{f}} D \quad \text { or } \quad \frac{1}{2} M(0.963 g L)\left(\frac{m}{M}\right)^{2}=(0.2 M g)(D) $$ from which \(D=2.4(m / M)^{2} L\).

Short Answer

Expert verified
The block slides a distance \(D = 2.4 \left(\frac{m}{M}\right)^2 L\)."

Step by step solution

01

Calculate Initial Speed at Bottom

The pendulum ball is released and falls through a height of \(0.201L\). Using conservation of energy, the speed of the ball at the bottom just before it hits the block is given by \(v = \sqrt{2g(0.201L)}\).
02

Calculate Rebound Speed

After rebounding, the ball reaches a height of \(0.0603L\). The speed when it starts rebounding upwards is \(v' = \sqrt{2g(0.0603L)}\).
03

Apply Conservation of Momentum

By conservation of momentum, \(m \sqrt{2g(0.201L)} = -m \sqrt{2g(0.0603L)} + MV\). Solve this for \(V\), the velocity of the block just after the collision, to find: \(V = \frac{m}{M} 0.981 \sqrt{gL}\).
04

Use Work-Energy Principle for the Block

The kinetic energy of the block is converted into work done against friction over distance \(D\). Thus, \(\frac{1}{2}M V^2 = F_f D\), where friction force \(F_f = 0.2Mg\).
05

Solve for Distance D

Substitute \(V\) from Step 3 into the equation from Step 4 and solve for \(D\): \(D = \frac{1}{2} \cdot M \left(\frac{m}{M} 0.981 \sqrt{gL}\right)^2 = 0.2MgD\). Simplifying, we find \(D = 2.4 \left(\frac{m}{M}\right)^2 L\).

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Key Concepts

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

Conservation of Energy
In the realm of physics, the principle of conservation of energy plays a pivotal role in understanding various mechanical systems. This principle states that energy within a closed system remains constant; it can neither be created nor destroyed but only transformed from one form to another.

In the pendulum exercise, the conservation of energy helps us determine the speeds of the ball at different points in its motion. Initially, the pendulum ball is at rest at a height of \(0.201L\) and thus, possesses potential energy (\(mgh\)). As the ball descends, this potential energy is completely converted into kinetic energy (\(\frac{1}{2} mv^2\)) at the lowest point. The equation \(mgh = \frac{1}{2} mv^2\) allows us to calculate the ball's speed at the bottom.

After the collision, some of the kinetic energy is converted back to potential energy, enabling us to determine how high the ball rebounds.
Kinematics
Kinematics deals with the motion of objects without looking at the forces that cause this motion. It plays an essential role in solving this problem, particularly in understanding how the pendulum behaves.

We utilize kinematics to analyze the movement of the ball as it swings down from a height of \(0.201L\) to its lowest point and back up following the rebound. Utilizing equations of motion permits us to determine variables such as speed, initial velocity, and distance traveled, which are critical in solving the exercise.

For instance, the approach of using the equation \( v = \sqrt{2gh} \) demonstrates how we can directly link the ball鈥檚 speed to its height, thanks to its kinematic characteristics.
Friction
Friction is an all-important force that acts to oppose the relative motion of two surfaces in contact. In this exercise, it affects the sliding block after the collision.

As the block glides to a stop, the friction force, given by \(0.2Mg\), performs work equal to the initial kinetic energy of the block. The equation \(\frac{1}{2}MV^2 = F_fD\) shows how the friction force does work over the distance \(D\) to bring the block to a halt.

Understanding friction is key to determining how far the block slides, as it ensures energy conservation principles apply even though the block's kinetic energy transforms entirely into thermal energy due to friction.
Pendulum Motion
Pendulum motion is a common example of periodic motion, occurring due to the gravitational force acting on the swinging ball. It offers insight into energy transformation processes through its cyclical swings.

Initially, the pendulum is at rest at a height before it swings down, converting potential energy into kinetic energy. Upon reversing its motion post-collision, it converts some of its kinetic energy back into potential energy until it finally comes to rest at its highest rebound point of \(0.0603L\).

A key property of pendulum motion is its predictable nature, allowing us to make exact calculations regarding any variations in speed, height, and energy, when factors like collision-induced rebounds and friction come into play.

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Most popular questions from this chapter

A \(1.0\) -kg ball moving at \(12 \mathrm{~m} / \mathrm{s}\) collides head-on with a \(2.0-\mathrm{kg}\) ball moving in the opposite direction at \(24 \mathrm{~m} / \mathrm{s}\). Determine the motion of each after impact if \((a) e=2 / 3,(b)\) the balls stick together, and (c) the collision is perfectly elastic. In all three cases the collision occurs along a straight line, and momentum is conserved. Hence, $$ \begin{aligned} \text { Momentum before } &=\text { Momentum after } \\ (1.0 \mathrm{~kg})(12 \mathrm{~m} / \mathrm{s})+(2.0 \mathrm{~kg})(-24 \mathrm{~m} / \mathrm{s}) &=(1.0 \mathrm{~kg}) v_{1}+(2.0 \mathrm{~kg}) v_{2} \end{aligned} $$ which becomes $$ -36 \mathrm{~m} / \mathrm{s}=v_{1}+2 v_{2} $$ (a) When \(e=2 / 3\), $$ e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \quad \text { becomes } \quad \frac{2}{3}=\frac{v_{2}-v_{1}}{(12 \mathrm{~m} / \mathrm{s})-(-24 \mathrm{~m} / \mathrm{s})} $$ from which \(24 \mathrm{~m} / \mathrm{s}=v_{2}-v_{1}\). Combining this with the momentum equation found above gives \(v_{2}=-4.0 \mathrm{~m} / \mathrm{s}\) and \(v_{1}=-28 \mathrm{~m} / \mathrm{s}\) (b) In this case \(v_{1}=v_{2}=v\), and so the momentum equation becomes $$ 3 v=-36 \mathrm{~m} / \mathrm{s} \quad \text { or } \quad v=-12 \mathrm{~m} / \mathrm{s} $$ (c) Here \(e=1\), and $$ e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \quad \text { becomes } \quad 1=\frac{v_{2}-v_{1}}{(12 \mathrm{~m} / \mathrm{s})-(-24 \mathrm{~m} / \mathrm{s})} $$ from which \(v_{2}-v_{1}=36 \mathrm{~m} / \mathrm{s}\). Adding this to the momentum equation yields \(v_{2}=0\). Using this value for \(v_{2}\) then leads to \(v_{1}=-36 \mathrm{~m} / \mathrm{s}\).

A \(7.00-\mathrm{g}\) bullet moving horizontally at \(200 \mathrm{~m} / \mathrm{s}\) strikes and passes through a 150 -g tin can sitting on a post. Just after impact, the can has a horizontal speed of \(180 \mathrm{~cm} / \mathrm{s}\). What was the bullet's speed after leaving the can?

A system consists of the following masses in the \(x y\) -plane: \(4.0 \mathrm{~kg}\) at coordinates \((x=0, y=5.0 \mathrm{~m})\), \(7.0 \mathrm{~kg}\) at \((3.0 \mathrm{~m}, 8.0 \mathrm{~m})\), and \(5.0 \mathrm{~kg}\) at \((-3.0 \mathrm{~m},-6.0 \mathrm{~m})\). Find the position of its center of mass. $$ \begin{array}{l} x_{\mathrm{cm}}=\frac{\sum x_{i} m_{i}}{\sum m_{i}}=\frac{(0)(4.0 \mathrm{~kg})+(3.0 \mathrm{~m})(7.0 \mathrm{~kg})+(-3.0 \mathrm{~m})(5.0 \mathrm{~kg})}{(4.0+7.0+5.0) \mathrm{kg}}=0.38 \mathrm{~m} \\ y_{\mathrm{cm}}=\frac{\sum y_{i} m_{i}}{\sum m_{i}}=\frac{(5.0 \mathrm{~m})(4.0 \mathrm{~kg})+(8.0 \mathrm{~m})(7.0 \mathrm{~kg})+(-6.0 \mathrm{~m})(5.0 \mathrm{~kg})}{16 \mathrm{~kg}}=2.9 \mathrm{~m} \end{array} $$ and \(z_{\mathrm{cm}}=0\). These distances are, of course, measured from the origin \((0,0,0,\), .

A ball of mass \(m\) sits at the coordinate origin when it explodes into two pieces that shoot along the \(x\) -axis in opposite directions. When one of the pieces (which has mass \(0.270 \mathrm{~m}\) ) is at \(x=70 \mathrm{~cm}\), where is the other piece? [Hint: What happens to the mass center?]

What force is exerted on a stationary flat plate held perpendicular to a jet of water as shown in Fig. 8-5? The horizontal speed of the water is \(80 \mathrm{~cm} / \mathrm{s}\), and \(30 \mathrm{~mL}\) of the water hit the plate each second. Assume the water moves parallel to the plate after striking it. One milliliter (mL) of water has a mass of \(1.00 \mathrm{~g}\). This question deals with speed, mass, time, and force, and that suggests impulse-momentum and Newton's Second Law. The plate exerts an impulse on the water and changes its horizontal momentum. The water exerts a counterforce on the plate. Taking the direction to the right as positive, \((\text { Impulse })_{x}=\) Change in \(x\) -directed momentum $$ F_{x} \Delta t=\left(m v_{x}\right)_{\text {final }}-\left(m v_{x}\right)_{\text {initial }} $$ Let \(t\) be \(1.00 \mathrm{~s}\) so that \(m\) will be the mass that strikes in \(1.00 \mathrm{~s}\), namely \(30 \mathrm{~g}\). Then the above equation becomes $$ F_{x}(1.00 \mathrm{~s})=(0.030 \mathrm{~kg})(0 \mathrm{~m} / \mathrm{s})-(0.030 \mathrm{~kg})(0.80 \mathrm{~m} / \mathrm{s}) $$ from which \(F_{x}=-0.024 \mathrm{~N}\). This is the force exerted by the plate on the water. The law of action and reaction tells us that the jet exerts an equal but opposite force on the plate.
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