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Chapter 8: Problem 72

A small wooden block with mass \(0.800 \mathrm{~kg}\) is suspended from the lower end of a light cord that is \(1.60 \mathrm{~m}\) long. The block is initially at rest. A bullet with mass \(12.0 \mathrm{~g}\) is fired at the block with a horizontal velocity \(c_{0}\) - The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of \(0.800 \mathrm{~m}\), the tension in the cord is \(4.80 \mathrm{~N}\). What was the initial speed \(t_{0}\) of the bullct?

Short Answer

Expert verified
The initial speed of bullet is \(270 \mathrm{~m/s}\).

Step by step solution

01

Finding the total mass

Find the total mass \(m_{total}\) of the block and the bullet system after the collision. Convert the bullet's mass from grams to kilograms and add it to the block's mass: \(m_{total} = 0.800 \mathrm{~kg} + 12.0 \mathrm{~g} = 0.812 \mathrm{~kg}\).
02

Calculating the speed of the pendulum after collision

Find the speed \(v_{after}\) of the combined block and bullet system after the collision using the principle of conservation of energy, where the potential energy at the highest point \(m_{total} \cdot g \cdot h\) equals the kinetic energy of the pendulum at the bottom, \(\frac{1}{2} \cdot m_{total} \cdot v_{after}^{2}\). Solving for \(v_{after}\) gives, \( v_{after} = \sqrt{2 \cdot g \cdot h} = \sqrt{2 \cdot 9.8 \mathrm{~m/s^2} \cdot 0.800 \mathrm{m}} = 4.0 \mathrm{~m/s}\).
03

Applying momentum conservation

Apply the principle of conservation of momentum. Initial momentum before the bullet strikes the block is the product of the bullet鈥檚 mass and initial speed \(m_{bullet} \cdot v_0\). After the collision, the total momentum is \(m_{total} \cdot v_{after}\). Thus, \(m_{bullet} \cdot v_0 = m_{total} \cdot v_{after}\). Solving for \(v_0\) gives, \(v_0 = \frac{m_{total} \cdot v_{after}}{m_{bullet}} = \frac{0.812 \mathrm{~kg} \cdot 4.0 \mathrm{~m/s}}{12.0 \mathrm{~g}} = 270 \mathrm{~m/s}\).

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Key Concepts

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

Kinetic and Potential Energy
Understanding the interplay between kinetic and potential energy is essential to solving problems like the one with the wooden block and bullet. Kinetic energy is the energy an object possesses due to its motion. It can be calculated using the formula \( KE = \frac{1}{2} mv^2 \), where \(m\) is the mass of the object and \(v\) is its velocity.

Potential energy, on the other hand, is stored energy based on an object's position or configuration. Gravitational potential energy, represented as \( PE = mgh \), depends on the mass of the object (\(m\)), the acceleration due to gravity (\(g\)), and the height (\(h\)) above a reference point.

In the provided exercise, the bullet鈥檚 kinetic energy just before collision transforms into both kinetic energy of the combined system (bullet plus wooden block) and potential energy as the block-bullet system rises to its highest point post-collision.
Collisions in Physics
Collisions in physics embody the principle of momentum conservation. The total momentum of a closed system remains constant before and after a collision.

In an inelastic collision, such as the one described between the bullet and the wooden block, the objects stick together after impact. The initial momentum, coming exclusively from the bullet, is transferred to the combined mass system of both the block and the bullet. According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision, which can be mathematically expressed as \( m_1v_1 + m_2v_2 = (m_1 + m_2)v_{after} \), where subscript 1 refers to the bullet and subscript 2 to the block.

Calculating the speed of the bullet involves using this principle. The initial speed \(v_0\) of the bullet is found by equating the momentum of the bullet before the collision to the momentum of the combined block and bullet system afterward.
Pendulum Motion
Pendulum motion plays a crucial role in this problem, as the block swings up to a certain height after the collision. A pendulum's motion is governed by gravitational forces, which convert potential energy back and forth into kinetic energy. When the block with the bullet embedded reaches its highest point, its kinetic energy is temporarily zero and all the energy in the system is potential.

The highest point reached by the pendulum corresponds to the maximum gravitational potential energy, which can be equated to the kinetic energy of the system just after the collision. This allows us to calculate the velocity of the block-bullet system post-collision, and ultimately reveal the initial velocity of the bullet by applying momentum conservation principles.

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Most popular questions from this chapter

Two ice skaters, Daniel (mass \(65.0 \mathrm{~kg}\) ) and Rebecca (mass \(45.0 \mathrm{~kg}\) ) are practicing. Danicl stops to tic his shoclace and, while at rest, is struck by Rebecca, who is moving at \(13.0 \mathrm{~m} / \mathrm{s}\) before she collides with him. After the collision, Rchecca has a velocity of magnitude \(8.00 \mathrm{~m} / \mathrm{s}\) at an angle of \(53.1^{\circ}\) from her initial direction, Both skaters move on the fric. tionless, horizontal surface of the rink. (a) What are the magnitude and direction of Danicl 's velocity after the collision? (b) What is the change in total kinetic energy of the two skaters as a result of the collision?

Two carts of cqual mass are on a horizontal, frictionless air track. Initially cart \(A\) is moving toward stationary cart \(B\) with a speed of \(v_{A}\). The carts undergo an inelastic collision, and after the collision the total kinetic energy of the two carts is one-half their initial total kinctic energy before the collision. What is the speed of cach cart after the collision?

A packing crute with mass \(80.0 \mathrm{~kg}\) is at rest on a horicontal. frictionlcss surface. At \(t=0\) a net horizontal force in the \(+x\) -dircction is applied to the crate. The force has a constant value of \(80.0 \mathrm{~N}\) for \(12.0 \mathrm{~s}\) and then decrcases lincarly with time so it becomes \(7 \mathrm{cro}\) after an ad. Jitional \(6.00 \mathrm{~s}\). What is the final speed of the crate. \(18.0 \mathrm{~s}\) after the force was first applied?

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.

In a shipping company distribution center, an open cart of } mass \(50.0 \mathrm{~kg}\) is rolling to the left at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\) (lig. \(\mathrm{P} 8.87\) ). Ignore friction between the cart and the floor. \(A 15.0 \mathrm{~kg}\) 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 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 packagc just bcfore it lands in the cart and (b) the final speed of the cart?
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