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Chapter 8: Problem 50

A 750 -kg cannon fires a 15 -kg projectile with a speed of \(250 \mathrm{~m} / \mathrm{s}\) with respect to the muzzle. The cannon is on wheels and can recoil with negligible friction. Just after the cannon fires the projectile, what is the speed of the projectile with respect to the ground?

Short Answer

Expert verified
Answer: The final velocity of the projectile with respect to the ground is 255 m/s.

Step by step solution

01

Understanding the conservation of momentum

The total momentum of a closed system is conserved when no external forces act upon it. In this case, we have a projectile and a cannon on wheels with negligible friction, so we can consider this a closed system. The conservation of momentum states that the total momentum before the interaction equals the total momentum after the interaction.
02

Define known quantities

We are given the following information: Mass of cannon, \(m_c = 750 \mathrm{~kg}\) Mass of projectile, \(m_p = 15 \mathrm{~kg}\) Speed of projectile with respect to the muzzle, \(v_p = 250 \mathrm{~m/s}\)
03

Calculate the total momentum before firing

Initially, the projectile and cannon are both at rest, so their total initial momentum is zero. Mathematically, this can be written as: \(P_{initial} = m_c \times 0 + m_p \times 0 = 0\)
04

Calculate the total momentum after firing

After firing the projectile, the cannon recoils, and the projectile moves forward. Let's assume the projectile moves in the positive direction and the cannon recoils in the negative direction. The total momentum after firing can be given by: \(P_{final} = m_c \times v_c + m_p \times v_p\) Here, \(v_c\) is the velocity of the cannon, which we need to find.
05

Apply conservation of momentum

Using the conservation of momentum, we know that the initial momentum must equal the final momentum: \(P_{initial} = P_{final}\) Substituting the values from steps 3 and 4, we get: \(0 = m_c \times v_c + m_p \times v_p\) Now, we need to isolate \(v_c\): \(v_c = -\frac{m_p \times v_p}{m_c}\) Substitute the given values: \(v_c = -\frac{15 \mathrm{~kg} \times 250 \mathrm{~m/s}}{750 \mathrm{~kg}} = -5 \mathrm{~m/s}\) Since the result is negative, the cannon moves in the opposite direction of the projectile, as expected.
06

Calculate the velocity of the projectile with respect to the ground

We've found the velocity of the projectile (\(v_p\)) with respect to the muzzle and the velocity of the cannon (\(v_c\)). To find the velocity of the projectile relative to the ground, we need to add the velocities of the projectile and cannon - remember that the cannon is recoiling, so the projectile velocity must be added to the absolute value of the cannon's velocity: \(v_{pg} = 250 \mathrm{~m/s} + |-5\mathrm{~m/s}| = 255 \mathrm{~m/s}\) The speed of the projectile with respect to the ground is \(255 \mathrm{~m/s}\).

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Most popular questions from this chapter

A student with a mass of \(40.0 \mathrm{~kg}\) can throw a \(5.00-\mathrm{kg}\) ball with a relative speed of \(10.0 \mathrm{~m} / \mathrm{s}\). The student is standing at rest on a cart of mass \(10.0 \mathrm{~kg}\) that can move without friction. If the student throws the ball horizontally, what will the velocity of the ball with respect to the ground be?

Two point masses are located in the same plane. The distance from mass 1 to the center of mass is \(3.0 \mathrm{~m} .\) The distance from mass 2 to the center of mass is \(1.0 \mathrm{~m} .\) What is \(m_{1} / m_{2},\) the ratio of mass 1 to mass \(2 ?\) a) \(3 / 4\) c) \(4 / 7\) e) \(1 / 3\) b) \(4 / 3\) d) \(7 / 4\) f) \(3 / 1\)

An astronaut is performing a space walk outside the International Space Station. The total mass of the astronaut with her space suit and all her gear is \(115 \mathrm{~kg} .\) A small leak develops in her propulsion system and \(7.00 \mathrm{~g}\) of gas are ejected each second into space with a speed of \(800 \mathrm{~m} / \mathrm{s}\). She notices the leak 6.00 s after it starts. How much will the gas leak have caused her to move from her original location in space by that time?

Young acrobats are standing still on a circular horizontal platform suspended at the center. The origin of the two-dimensional Cartesian coordinate system is assumed to be at the center of the platform. A 30.0 -kg acrobat is located at \((3.00 \mathrm{~m}, 4.00 \mathrm{~m})\), and a 40 - \(\mathrm{kg}\) acrobat is located at \((-2.00 \mathrm{~m}\) \(-2.00 \mathrm{~m})\). Assuming that the acrobats stand still in their positions, where must a 20.0 -kg acrobat be located so that the center of mass of the system consisting of the three acrobats is at the origin and the platform is balanced?

Two masses, \(m_{1}=2.0 \mathrm{~kg}\) and \(m_{2}=3.0 \mathrm{~kg}\), are moving in the \(x y\) -plane. The velocity of their center of mass and the velocity of mass 1 relative to mass 2 are given by the vectors \(v_{\mathrm{cm}}=(-1.0,+2.4) \mathrm{m} / \mathrm{s}\) and \(v_{\mathrm{rel}}=(+5.0,+1.0) \mathrm{m} / \mathrm{s} .\) Determine a) the total momentum of the system b) the momentum of mass 1 , and c) the momentum of mass 2 .
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