A block with mass$$

m$1=0\mathrm{.}450$kg

$$is released from rest on a frictionless track at a distance$$

h$1=2\mathrm{.}45$m

$$above the top of a table. It then collides elastically with an object having mass$$

m$2=0\mathrm{.}900$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$$h$1.$ The height of the table is$$h$2.$ A block labeled$$m$1$rests on the top of the wedge. A block labeled$$m$2$rests at the very bottom of the slope of the wedge, at the edge of the table. A dashed parabolic line shows the path$$m$2$would take after being knocked off the table. The path ends a horizontal distance$$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.$)$

v$1=$m$/$sv$2=$m$/$s

$($b$)$

How high up the track does the$0\mathrm{.}450$kg$$object travel back after the collision?

$$m

$($c$)$

How far away from the bottom of the table does the$0\mathrm{.}900$kg$$object land, given that the height of the table is$$

h$2=1\mathrm{.}80$m$?$

$$m

$($d$)$

How far away from the bottom of the table does the$0\mathrm{.}450$kg$$object eventually land?

$$m