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Chapter 6: Problem 23

A 5.18-g bullet moving at \(672 \mathrm{~m} / \mathrm{s}\) strikes a \(715-\mathrm{g}\) wooden block at rest on a frictionless surface. The bullet emerges with its speed reduced to \(428 \mathrm{~m} / \mathrm{s}\). Find the resulting speed of the block.

Short Answer

Expert verified
The resulting speed of the block can be found by solving the above equation which is obtained using the conservation of momentum principle.

Step by step solution

01

Understand The Problem

A bullet passes through a wooden block, the bullet slows down as a result. As the bullet and wooden block system is an isolated system, we can apply the law of conservation of momentum.
02

Initial Momentum Calculation

Firstly, calculate the initial momentum in the system. In this system, before collision, only bullet is moving so the initial momentum will be the momentum of the bullet. In momentum formula, momentum = mass * velocity. For the bullet, mass( \(m_{1}\) )= 5.18 g = 0.00518 kg (Convert grams into kilograms making sure SI units are used) and the velocity ( \(v_{1i}\)) = 672 m/s, so the initial momentum ( \(P_{i}\)) = \(m_{1} * v_{1i} \).
03

Final Momentum Calculation

The final momentum is the sum of the momentum of the bullet after passing through the block and the momentum of the wooden block. For the bullet its mass ( \(m_{1}\) )= 0.00518 kg and its final velocity ( \(v_{1f}\)) = 428 m/s. Therefore, the final momentum of the bullet equals \(m_{1} * v_{1f}\). The block is now moving at a velocity( \(v_{2f}\) ) which we are asked to find, and has a mass ( \(m_{2}\) ) = 715 g = 0.715 kg. Hence the final momentum of the block equals \(m_{2} * v_{2f}\). So, the final total momentum = momentum of bullet after passing + momentum of block = \(m_{1} * v_{1f} + m_{2} * v_{2f}\)
04

Solve For Unknown

Now, use conservation of momentum to equate the initial and final momentum and solve for the unknown velocity. Therefore, \(m_{1} * v_{1i} = m_{1} * v_{1f} + m_{2} * v_{2f}\)

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Most popular questions from this chapter

In the laboratory, a particle of mass \(3.16 \mathrm{~kg}\) moving at \(15.6 \mathrm{~m} / \mathrm{s}\) to the left collides head-on with a particle of mass \(2.84 \mathrm{~kg}\) moving at \(12.2 \mathrm{~m} / \mathrm{s}\) to the right. Find the velocity of the center of mass of the system of two particles after the collision.

A space vehicle is traveling at \(3860 \mathrm{~km} / \mathrm{h}\) with respect to the Earth when the exhausted rocket motor is disengaged and sent backward with a speed of \(125 \mathrm{~km} / \mathrm{h}\) with respect to the command module. The mass of the motor is four times the mass of the module. What is the speed of the command module after the separation?

A golfer hits a golf ball, imparting to it an initial velocity of magnitude \(52.2 \mathrm{~m} / \mathrm{s}\) directed \(30^{\circ}\) above the horizontal. Assuming that the mass of the ball is \(46.0 \mathrm{~g}\) and the club and ball are in contact for \(1.20 \mathrm{~ms}\), find \((a)\) the impulse imparted to the ball, \((b)\) the impulse imparted to the club, and \((c)\) the average force exerted on the ball by the club.

After a totally inelastic collision, two objects of the same mass and initial speed are found to move away together at half their initial speed. Find the angle between the initial velocities of the objects.

A ball of mass \(m\) and speed \(v\) strikes a wall perpendicularly and rebounds with undiminished speed. ( \(a\) ) If the time of collision is \(\Delta t\), what is the average force exerted by the ball on the wall? (b) Evaluate this average force numerically for a rubber ball with mass \(140 \mathrm{~g}\) moving at \(7.8 \mathrm{~m} / \mathrm{s}\); the duration of the collision is \(3.9 \mathrm{~ms}\).
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