In bocce, the object of the game is to get your balls (each with mass \(M=1.00 \mathrm{~kg}\) ) as close as possible to the small white ball (the pallina, mass \(m=0.045 \mathrm{~kg}\) ). Your first throw positioned your ball \(2.00 \mathrm{~m}\) to the left of the pallina. If your next throw has a speed of \(v=1.00 \mathrm{~m} / \mathrm{s}\) and the coefficient of kinetic friction is \(\mu_{\mathrm{k}}=0.20\), what are the final distances of your two balls from the pallina in each of the following cases? a) You throw your ball from the left, hitting your first ball. b) You throw your ball from the right, hitting the pallina.

In this step, we need to understand the given problem and collect all the necessary information, which includes the mass of the balls, initial positions and velocities, and the coefficient of kinetic friction. Given data: - Mass of the first ball (M) = 1.00 kg - Mass of the pallina (m) = 0.045 kg - Initial position of your first ball = 2.00 m to the left of the pallina - The speed of your next throw (v) = 1.00 m/s - The coefficient of kinetic friction (μk) = 0.20

Before the collision, the momentum of all the objects involved must equal the total momentum after the collision. Using the conservation of momentum principle, we can write the equation for each case: a) When you throw your ball from the left and hit the first ball: Initial momentum = Final momentum M * v + M * 0 = M * v1 + M * v2 b) When you throw your ball from the right and hit the pallina: Initial momentum = Final momentum m * 0 + M * v = m * vp + M * vr Where: - v1 = The final velocity of your first ball in case a - v2 = The final velocity of your second ball in case a - vp = The final velocity of the pallina in case b - vr = The final velocity of your second ball in case b

Using Newton's second law (F = ma), we can calculate the final velocities of the balls and pallina, considering the friction forces acting on them: a) Velocity of first ball in case a: M * v1 = M * a1 => v1 = a1 Velocity of second ball in case a: M * v2 = M * a2 => v2 = a2 Friction forces: a1 = -μk * g, a2 = μk * g b) Velocity of pallina in case b: m * vp = m * ap => vp = ap Velocity of second ball in case b: M * vr = M * ar => vr = ar Friction forces: ap = μk * g, ar = -μk * g Where: - a1, a2 = acceleration of balls in case a due to kinetic friction - ap, ar = acceleration of pallina and ball in case b due to kinetic friction - g = acceleration due to gravity (approximately 9.8 m/s²)

Now, we'll find the final distances of your two balls from the pallina using kinematic equations: a) Distance traveled by first ball: d1 = (1/2) * a1 * t1^2 Distance traveled by second ball: d2 = (1/2) * a2 * t2^2 b) Distance traveled by pallina: dp = (1/2) * ap * tp^2 Distance traveled by second ball: dr = (1/2) * ar * tr^2 Where t1, t2, tp, and tr are the time intervals balls and pallina take to come to rest. To find the time intervals, we can use the equation vf = vi + a*t. Since the final velocities of all objects will be zero, the time intervals can be calculated as t = -vi/a. Finally, to determine the final distances of the balls and pallina from their initial positions, we can add or subtract the distances traveled by them in each case: a) Final distance of first ball from pallina: 2.00 m + d1 Final distance of second ball from pallina: 2.00 m - d2 b) Final distance of pallina from the initial position: dp Final distance of second ball from pallina: dr + dp Remember to calculate the values for a1, a2, ap, ar, t1, t2, tp, tr, d1, d2, dp, and dr using the given values and equations before calculating the final distances for each case.