Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.

W = 3.266 N
The mass of the meters stick is :

So, the mass of the meter stick is 0.333 kg.
Answer:
0.6kg
Explanation:
the unknown here is the mass of the second block
applying the law of the conservation of momentum
m₁v₁ + m₂v₂ = (m₁ + m₂) v₃
where m₁=mass of first block=2.2kg
m₂=mass of colliding block= ?
v₁= velocity of first block=1.2m/s
v₂=velocity of colliding block=4.0m/s
v₃= final velocity of combined block=1.8m/s
applying the formula above
(2.2 × 1.2) + (m₂ × 4) = (2.2 + m₂) × 1.8
2.64 + 4m₂ = 3.96 + 1.8m₂
collecting like terms
4m₂ - 1.8m₂ = 3.96 - 2.64
2.2m₂=1.32
divide both sides by 2.2
m₂= 0.6kg