Block \(A,\) with a mass of \(220 \mathrm{g},\) is traveling north on a frictionless surface with a speed of \(5.0 \mathrm{m} / \mathrm{s} .\) Block \(\mathrm{B}\) with a mass of \(300 \mathrm{g}\) travels west on the same surface until it collides with A. After the collision, the blocks move off together with a velocity of \(3.13 \mathrm{m} / \mathrm{s}\) at an angle of \(42.5^{\circ}\) to the north of west. What was \(\mathrm{B}\) 's speed just before the collision?

First, convert masses of block A and B to kilograms: \(220 \mathrm{g} = 0.22 \mathrm{kg}\) \(300 \mathrm{g} = 0.3 \mathrm{kg}\)

The final velocity is given in polar coordinates, so first, we need to convert it to cartesian coordinates by finding the x and y-components of the final momentum. Final velocity magnitude: \(V_f = 3.13 \mathrm{m} / \mathrm{s}\) Final velocity angle: \(θ = 42.5^\circ\) Mass of both blocks combined: \(m_A + m_B = 0.22 \mathrm{kg} + 0.3 \mathrm{kg} = 0.52 \mathrm{kg}\) Magnitude of the final momentum: \(P_f = m_{total} \cdot V_f = 0.52 \mathrm{kg} \cdot 3.13 \mathrm{m}/\mathrm{s} = 1.6276 \mathrm{kg}\,\mathrm{m}/\mathrm{s}\) We find the x and y components of the final momentum: \(P_{fx} = P_f * \cos{(42.5^\circ)} = 1.6276 * \cos{(42.5)} = 1.1969 \, \mathrm{kg}\,\mathrm{m}/\mathrm{s}\) \(P_{fy} = P_f * \sin{(42.5^\circ)} = 1.6276 * \sin{(42.5)} = 1.1028 \, \mathrm{kg}\,\mathrm{m}/\mathrm{s}\)

We know block A's initial speed and direction, so its initial momentum can be calculated directly. Since block B is moving west, its y-component of momentum is 0 before the collision. The initial momentum of block A: \(P_{Ai} = m_A \cdot v_{A} = 0.22 \mathrm{kg} \cdot 5.0 \mathrm{m}/\mathrm{s} = 1.1 \, \mathrm{kg}\,\mathrm{m}/\mathrm{s}\) Block A is moving north, so \(P_{Aix} = 0\), and \(P_{Aiy} = P_{Ai} = 1.1 \, \mathrm{kg}\,\mathrm{m}/\mathrm{s}\) The initial momentum of block B: \(P_{Bix}\) is unknown, and \(P_{Biy} = 0\)

Now, we can use the conservation of linear momentum in both x and y directions. In the x-direction: \(P_{Aix} + P_{Bix} = P_{fx}\) \(0 + P_{Bix} = 1.1969 \, \mathrm{kg}\,\mathrm{m}/\mathrm{s}\) So, \(P_{Bix} = 1.1969 \, \mathrm{kg}\,\mathrm{m}/\mathrm{s}\). In the y-direction: \(P_{Aiy} + P_{Biy} = P_{fy}\) \(1.1 \, \mathrm{kg}\,\mathrm{m}/\mathrm{s} + 0 = 1.1028 \, \mathrm{kg}\,\mathrm{m}/\mathrm{s}\) Since the initial and final values are close, it seems reasonable that the blocks don't lose any significant momentum in the y-direction.

Finally, we can find the initial speed of block B by dividing its initial momentum by its mass: \(v_{Bi} = \frac{P_{Bix}}{m_B} = \frac{1.1969 \, \mathrm{kg}\,\mathrm{m}/\mathrm{s}}{0.3 \, \mathrm{kg}} = 3.99 \, \mathrm{m}/\mathrm{s}\) Therefore, the initial speed of block B just before the collision was \(3.99 \, \mathrm{m}/\mathrm{s}\).