The question is incomplete. The mass of the object is 10 gram and travelling at a speed of 2 m/s.
Solution:
It is given that mass of object before explosion is,m = 10 g
Speed of object before explosion, v = 2 m/s
Let
be the masses of the three fragments.
Let
be the velocities of the three fragments.
Therefore, according to the law of conservation of momentum,


So the x- component of the velocity of the m2 fragment after the explosion is,

∴ 
I’m not really sure but I think it’s D type 1 lever
Answer:
Explanation:
On the Moon :----
1500 x 1.6 = 2400 m /s is initial velocity of bullet .
g = 1.6 m /s²
v = u - gt
0 = 2400 - 1.6 t
t = 1500 s
This is time of ascent
Time of decent will also be the same
Total time of flight = 2 x 1500 = 3000 s
On the Earth : ---
v = u - a₁ t
0 = u - a₁ x 18
u = 18a₁
v² = u² - 2 x a₁ x 2743.2
0 = (18a₁ )² - 2 x a₁ x 2743.2
a₁ = 16.93
For downward return
s = ut + 1/2 a₂ x t²
2743.2 = 0 + .5 x a₂ x 31²
a₂ = 5.7 m /s²
If d be the deceleration produced by air
g + d = 16.93 ( during upward journey )
g - d = 5.7
g = (16.93 + 5.7) / 2
= 11.315 m / s
d = 5.6 m /s²
So air is creating a deceleration of 5.6 m /s².