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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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