A small block of mass m1 is released from rest at the top of a curve-shaped, frictionless wedge which sits on a frictionless horizontal surface. The height of the wedge is h=5m. When the block leaves the wedge its velocity relative to the ground is measured to be 4 m/s to the right. If the mass of the block is doubled to become 2m1 what, what will its speed be when it leaves the wedge?

I think "m and M" might be clearer than "m1 and m2."

There are no external horizontal forces applied, so horizontal momentum is conserved. For the ramp of mass M,
p = 0 = m4.00m/s - MV
giving V a positive sign. So
V = 4m/s m / M

Since there is no friction, energy is also conserved, so the block's initial PE becomes KE in the two objects:
mgh = ½(m(4.00m/s)² + M(4m/s m/M)²)
m 9.8m/s² 5m = m8m²/s² + 8m²/s² m²/M → divide by m
49m²/s² = 8m²/s² + 8m²/s² m/M
m/M = 41m²/s² / 8m²/s² = 5.125
M = m / 5.125

For a block of doubled mass, our conservation of momentum equation becomes
p = 0 = 2mv - (m/5.125)V
which rearranges to
V = 10.25v

and so our conservation of energy equation becomes
2mgh = ½(2mv² + (m/5.125)(10.25v)²) = mv² + 10.25mv² → m cancels
2gh = 11.25v²
gh = 5.625v²
so
v² = gh / 5.625 = 49m²/s² / 5.625 = 8.71 m²/s²
v = 2.95 m/s ◄

Check:
for the block of mass m, V = 4m/s 5.125 = 20.5 m/s
for a total KE = ½(m4² + (m/5.125)(20.5)²) = 49m J → for m in kg

for the block of mass 2m, V = 10.25 2.95m/s = 30.25 m/s
and the total KE = ½(2m*(2.95)² + (m/5.125)(30.25)²) = 98m J → for m in kg

and we have in fact finished with twice the KE (which is good, since we finished with twice the starting PE).

Hope this helps!