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 original height h to the height at which the collision occurs.<br /><br />Let's denote the time it takes for the clay to fall as t. The distance traveled by the clay can be calculated using the equation:<br /><br />s = ut + 1/2gt^2<br /><br />where s is the distance, u is the initial velocity, g is the acceleration due to gravity, and t is the time.<br /><br />Since the clay is released from rest, u = 0. Substituting the given values, we have:<br /><br />s = 0 + 1/2(10)t^2<br />s = 5t^2<br /><br />The distance traveled by the clay is equal to the height h minus the height at which the collision occurs. Therefore, we have:<br /><br />h - s = h - 5t^2<br /><br />Solving for t, we get:<br /><br />t = √(h/5)<br /><br />Substituting the given value of h = 10 m, we have:<br /><br />t = √(10/5)<br />t = √2 ≈ 1.41 s<br /><br />Therefore, the time after the piece of clay is released at which the collision takes place is approximately 1.41 seconds.<br /><br />c. To determine the height above the ground at which the collision takes place, we need to find the height of the ball at the time t when the collision occurs.<br /><br />The height of the ball at time t can be calculated using the equation:<br /><br />h = ut + 1/2gt^2<br /><br />Since the ball is dropped from rest, u = 0. Substituting the given values, we have:<br /><br />h = 0 + 1/2(10)t^2<br />h = 5t^2<br /><br />Substituting the value of t = √2 ≈ 1.41 s, we have:<br /><br />h = 5(1.41)^2<br />h ≈ 9.95 m<br /><br />Therefore, the height above the ground at which the collision takes place is approximately 9.95 m.<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.<br /><br />Let's denote the speed of the ball immediately before the collision as v1 and the speed of the piece of clay as v2.<br /><br />Using the conservation of momentum, we have:<br /><br />4m * v1 + m * v2 = 4m * v1' + m * v2'<br /><br />where v1' and v2' are the speeds of the ball and the piece of clay immediately after the collision.<br /><br />Using the conservation of kinetic energy, we have:<br /><br />1/2 * 4m * v1^2 + 1/2 * m * v2^2 = 1/2 * 4m * v1'^2 + 1/2 * m * v2'^2<br /><br />Solving these two equations simultaneously, we can find the values of v1 and v2.<br /><br />e. If the ball and the clay stick together on impact, their combined mass will be 5m. Using the conservation of momentum, we can find their velocity immediately after the collision:<br /><br />(4m * v1 + m * v2) / (4m + m) = v_f<br /><br />where v_f is the final velocity of the combined system.