A catapult on a level field tosses a 3 -kg stone a horizontal distance of \(100 \mathrm{~m}\). A second 3 -kg stone tossed in an identical fashion breaks apart in the air into 2 pieces, one with a mass of \(1 \mathrm{~kg}\) and one with a mass of \(2 \mathrm{~kg} .\) Both of the pieces hit the ground at the same time. If the 1 -kg piece lands a distance of \(180 \mathrm{~m}\) away from the catapult, how far away from the catapult does the 2 -kg piece land? Ignore air resistance. a) \(20 \mathrm{~m}\) c) \(100 \mathrm{~m}\) e) \(180 \mathrm{~m}\) b) \(60 \mathrm{~m}\) d) \(120 \mathrm{~m}\)

The given values are: 1. Mass of the first stone: 3 kg 2. Mass of the broken pieces: 1 kg and 2 kg 3. Landing distance for the unbroken stone: 100 m 4. Landing distance for the 1 kg piece: 180 m Note that the final velocity of the 2 kg piece is the same in the x and y components as the velocity of the 1 kg piece, since they both hit the ground at the same time.

Since momentum is conserved, we can write: Initial momentum (before breaking) = Final momentum (after breaking) Let's denote the velocities of the two broken pieces as: \(v_1\) - velocity of 1 kg piece \(v_2\) - velocity of 2 kg piece The initial momentum (3 kg stone) can be found by using the range equation for projectile motion: \(v_{initial} = \frac{100 \mathrm{~m} \cdot g}{v_{vertical}}\) The vertical component of velocity will cancel out due to the stone's parabolic nature. Therefore: \(M_initial \cdot v_{initial}=M_1 \cdot v_1 + M_2 \cdot v_2\)

Since both broken pieces hit the ground at the same time, they have the same horizontal velocity. Let's denote the horizontal velocity as \(v_{horizontal}\). We can find \(v_1\) by using the range equation again: \(v_1 = \frac{180 \mathrm{~m} \cdot g}{v_{vertical}}\) Now we can plug this back into the momentum conservation equation: \(3 \cdot v_{initial} = 1 \cdot v_1 + 2 \cdot v_2\) Using the range equations: \(3 \cdot \frac{100 \mathrm{~m} \cdot g}{v_{vertical}} = 1 \cdot \frac{180 \mathrm{~m} \cdot g}{v_{vertical}} + 2 \cdot v_2\) Cancel out the \(g\) and \(v_{vertical}\), and we get: \(3 \cdot 100 = 180 + 2 \cdot v_2\)

After calculating the velocity of the 2 kg piece from the above equation, we can use the range equation again to find the distance it travels: \(v_2 = \frac{3\cdot100 - 180}{2} = 60\) Finally, using the range equation: \(distance_{2kg} = \frac{v_2 \cdot v_{vertical}}{g} = \frac{60 \mathrm{~m} \cdot v_{vertical}}{g}\) Since the proportion is consistent, we can set this equal to: \(100 \mathrm{~m} = \frac{v_{initial} \cdot v_{vertical}}{g}\) Solve for \(distance_{2kg}\): \(\frac{60 \mathrm{~m} \cdot v_{vertical}}{g} = \frac{v_{initial} \cdot v_{vertical}}{100} \Rightarrow distance_{2kg} = 60 \mathrm{~m}\) The 2 kg piece lands at a distance of 60 meters from the catapult. The correct answer is (b) \(60 \mathrm{~m}\).