A body weighing 1 kg falls from a height of 2 m with an acceleration of 8 m / s2.

A body weighing 1 kg falls from a height of 2 m with an acceleration of 8 m / s2. What is the momentum of the body at the moment of impact on the Earth?

Given:
m = 0.2 kg;
h = 1 m;
a = 8 m / s².
Find: delta p.
Decision:
Let’s write two momentum formulas:
p1 = mV1,
p2 = mV2.
To find the delta p, one must subtract one impulse from the other.
∆p = p2-p1.
We substitute instead of impulses, the right-hand sides of the equations.
∆p = mV2-mV1
Due to the fact that V1 = 0, the formula will be like this:
∆p = mV2.
Next, we write the distance formula:
S = at² / 2.
We need to find t, so we also write a formula for it:
t = √ (2S / a).
Substitute
t = √ (2/8).
Turns out that
t = 0.5 s.
Then we find the acceleration using the formula:
a = V2-V0 / t.
Again we take into account that V0 = 0. This means that the formula takes this form:
a = V2 / t.
Let us express V2 from it.
V2 = at.
Substitute the known values:
V2 = 0.8 * 0.5 s = 0.4 m / s.
We can return to the delta p formula, since we now know all the values. We substitute them and find the answer.
∆p = 0, 4 * 0, 2 = 0, 08 kg * m / s.
Answer: 0.08