A block with mass m1 = 0.400 kg is released from rest on a frictionless track at a distance h1 = 3.00 m above the top of a table. It then collides elastically with an object having mass m2 = 0.800 kg that is initially at rest on the table, as shown in the figure below.A curved wedge rests on a table. The wedge has the appearance of a hill sliced vertically in half: the surface at the top is horizontal, and then it gradually increases the magnitude of its slope down to the right before turning sharply down, and then decreasing the magnitude of its slope until it is almost horizontal. The vertical distance between the surface of the table and the top of the wedge is h1. The height of the table is h2. A block labeled m1 rests on the top of the wedge. A block labeled m2 rests at the very bottom of the slope of the wedge, at the edge of the table. A dashed parabolic line shows the path m2 would take after being knocked off the table. The path ends a horizontal distance x from the edge of the table.(a) Determine the velocities of the two objects just after the collision. (Assume the positive direction is to the right. Indicate the direction with the signs of your answers.)v1 = m/sv2 = m/s(b) How high up the track does the 0.400-kg object travel back after the collision? m(c) How far away from the bottom of the table does the 0.800-kg object land, given that the height of the table is h2 = 1.95 m? m(d) How far away from the bottom of the table does the 0.400-kg object eventually land? m

(a) To determine the velocities of the two objects just after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.

The conservation of momentum before and after the collision is given by:

m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final

Given that both objects are initially at rest, the equation simplifies to:

m1 * v1_final = m2 * v2_final ----(1)

The conservation of kinetic energy before and after the collision is given by:

1/2 * m1 * (v1_initial)^2 + 1/2 * m2 * (v2_initial)^2 = 1/2 * m1 * (v1_final)^2 + 1/2 * m2 * (v2_final)^2

Given that both objects are initially at rest, and the initial velocity of m1 is sqrt(2gh1), the equation simplifies to:

1/2 * m1 * (sqrt(2gh1))^2 = 1/2 * m1 * (v1_final)^2 + 1/2 * m2 * (v2_final)^2

which simplifies to:

m1 * g * h1 = 1/2 * m1 * (v1_final)^2 + 1/2 * m2 * (v2_final)^2 ----(2)

Solving equations (1) and (2) simultaneously will give the final velocities v1_final and v2_final.

(b) The 0.400-kg object will travel back up the track to a height that can be determined by using the conservation of energy. The kinetic energy just after the collision will be converted into potential energy at the highest point of its trajectory. Therefore:

1/2 * m1 * (v1_final)^2 = m1 * g * h

Solving for h will give the height.

(c) The 0.800-kg object will land a distance away from the bottom of the table that can be determined by using the equations of projectile motion. The horizontal distance (x) it travels can be found using the equation:

x = v2_final * t

where t is the time it takes for the object to hit the ground. This can be found using the equation:

t = sqrt(2*h2/g)

(d) The 0.400-kg object will land a distance away from the bottom of the table that can be determined by using the equations of projectile motion, similar to part (c). The only difference is that the initial velocity will be the velocity of the 0.400-kg object just after the collision, and the height will be the height to which it travels back up the track.