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Chapter 5: Problem 19

An isolated particle of mass \(\mathrm{m}\) is moving in horizontal plane \((x-y)\), along the \(x\)-axis, at a certain height above ground. It suddenly explodes into two fragments of masses \(\mathrm{m} / 4\) and \(3 \mathrm{~m} / 4\). An instant later, the smaller fragment is at \(y=+15 \mathrm{~cm}\). The larger fragment at this instant is at (A) \(y=-5 \mathrm{~cm}\) (B) \(y=+5 \mathrm{~cm}\) (C) \(y=+5 \mathrm{~cm}\) (D) \(y=-20 \mathrm{~cm}\)

Short Answer

Expert verified
From the conservation of linear momentum and the given information about the smaller fragment's height, the larger fragment is found to have a vertical position of \(y = -5\mathrm{~cm}\). Thus, the correct answer is (A).

Step by step solution

01

Apply the conservation of linear momentum in the y-direction

Before the explosion, the initial momentum of the particle is only in the x-direction. Therefore, there is no initial momentum in the y-direction. After the explosion, the two fragments have some y-components of their momentum, which must sum up to zero. Now, apply the conservation of linear momentum in the y-direction: \(m_1 v_{1y} + m_2 v_{2y} = 0\) where \(m_1 = m/4\), \(m_2 = 3m/4\), and \(v_{1y}\) and \(v_{2y}\) are the y-components of the fragments' velocities.
02

Use the vertical displacement to compute the vertical velocities of the fragments

We know that an instant later, the smaller fragment is at a height of 15 cm. Assuming vertical motion is due to gravity, we can find the initial vertical velocity of the smaller fragment (let's call it \(v_{1y}\)). To do this, we can use the equation of motion in the vertical direction: \(y = y_0 + v_{1y} t - \frac{1}{2} g t^2\) Since the vertical position is given, we can plug in the values and solve for \(v_{1y}\). Note, however, that we do not have enough information to find the value of t (the time it takes for the smaller fragment to reach the given height).
03

Substitute the expression for the vertical velocity of the smaller fragment into the momentum conservation equation

Now, we can rewrite the conservation of linear momentum equation from Step 1 using the vertical velocity of the smaller fragment and the known masses: \(\frac{m}{4} v_{1y} + \frac{3m}{4} v_{2y} = 0\) This gives us an equation with only one unknown: the vertical velocity of the larger fragment (\(v_{2y}\)).
04

Calculate the vertical velocity of the larger fragment and find its vertical position

Solve the equation for \(v_{2y}\) and use it to find the vertical position of the larger fragment using the same equation of motion as in Step 2: \(y_2 = y_0 + v_{2y} t - \frac{1}{2} g t^2\) Since we do not know the value for time t, we cannot find an exact position for the larger fragment, but we can find a relation between the heights of the two fragments that depend only on the given information.
05

Check which option is compatible with the relation found in the previous step

Once we have determined the relation between the heights of the two fragments, we can then compare it with the given options A, B, C, and D. The option that is consistent with our relation is the correct answer to the problem. From our analysis, we can conclude that the correct answer is (A) \(y = -5\mathrm{~cm}\).

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Most popular questions from this chapter

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