A ball of mass 4m is dropped from rest from a height h=10.0m above the ground, as shown above. It undergoes a perfectly elastic collision with the ground and rebounds. At the instant that the ball rebounds, a small piece of clay of mass m is released from rest from the original height h directly above the ball as shown above on the right. The clay, which is descending , eventually collides with the ball, which is ascending Assume that g= 10m/s^2 air resistance is negligible , and the collision process takes negligible time. a. Determine the speed of the ball immediately before it hits the ground. b. Determine the time after the piece of clay is released at which the collision takes place. c. Determine the height above the ground at which the collision takes place. d. Determine the speeds of the ball and the piece of clay immediately before the collision. e. If the ball and the clay stick together on impact, what is the magnitude and direction of their velocity immediately after the collision?

a. To determine the speed of the ball immediately before it hits the ground, we can use the kinematic equation:<br /><br />v^2 = u^2 + 2gh<br /><br />where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and h is the height.<br /><br />Since the ball is dropped from rest, u = 0. Substituting the given values, we have:<br /><br />v^2 = 0 + 2(10)(10)<br />v^2 = 200<br />v = √200 ≈ 14.14 m/s<br /><br />Therefore, the speed of the ball immediately before it hits the ground is approximately 14.14 m/s.<br /><br />b. To determine the time after the piece of clay is released at which the collision takes place, we need to find the time it takes for the clay to fall from its initial height h to the height at which the collision occurs.<br /><br />Let's denote the height at which the collision occurs as h'. The time it takes for the clay to fall from h to h' can be calculated using the kinematic equation:<br /><br />h' = h - (1/2)gt^2<br /><br />where h' is the height at which the collision occurs, h is the initial height, g is the acceleration due to gravity, and t is the time.<br /><br />Since the collision occurs at the same time for both the ball and the clay, we can use the time calculated in part b as the time for the collision.<br /><br />c. To determine the height above the ground at which the collision takes place, we need to find the height h' at which the clay falls to meet the ball.<br /><br />Using the kinematic equation:<br /><br />h' = h - (1/2)gt^2<br /><br />where h' is the height at which the collision occurs, h is the initial height, g is the acceleration due to gravity, and t is the time.<br /><br />Substituting the given values, we have:<br /><br />h' = 10 - (1/2)(10)(t^2)<br /><br />d. To determine the speeds of the ball and the piece of clay immediately before the collision, we need to use the conservation of momentum and kinetic energy principles.<br /><br />Let's denote the speed of the ball immediately before the collision as v_ball and the speed of the clay as v_clay.<br /><br />Using the conservation of momentum:<br /><br />m_ball * v_ball = m_clay * v_clay<br /><br />Using the conservation of kinetic energy:<br /><br />(1/2) * m_ball * v_ball^2 = (1/2) * m_clay * v_clay^2<br /><br />e. If the ball and the clay stick together on impact, we can use the conservation of momentum to determine their velocity immediately after the collision.<br /><br />Let's denote the combined mass of the ball and clay as M and their velocity immediately after the collision as v_final.<br /><br />Using the conservation of momentum:<br /><br />M * v_final = m_ball * v_ball + m_clay * v_clay<br /><br />Since the ball and clay stick together, their velocities will be the same immediately after the collision. Therefore, we can simplify the equation:<br /><br />M * v_final = (m_ball + m_clay) * v_final<br /><br />Solving for v_final, we get:<br /><br />v_final = (m_ball + m_clay) / M * v_ball<br /><br />Since the ball and clay stick together, their combined mass M is equal to the sum of their individual masses:<br /><br />M = m_ball + m_clay<br /><br />Substituting this into the equation for v_final, we get:<br /><br />v_final = (m_ball + m_clay) / (m_ball + m_clay) * v_ball<br /><br />v_final = v_ball<br /><br />Therefore, the magnitude of their velocity immediately after the collision is equal to the speed of the ball immediately before the collision, and the direction is the same as the direction of the ball's velocity before the collision.