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A 50kg mass is sitting on a frictionless surface.An unknown constant force pushes the mass for 2 seconds until the mass reaches a velocity of 3m/s.
A.What is the initial momentum of the mass?
B.What is the final momentum of the mass?

Given:



Required:



Solution:​


Sagot :

Answer:

m = 50 kg

t = 3s

vi = 0 m/s

vf = 5 m/s

a. What is the final momentum of the mass?

Formula:

Δp = mΔv

Solution:

Δp = 50 (5 - 0)

Δp = 325

Δp = 300 m/s

The formula Δp = mΔv is the same or can also be written as Δp = m (vf-vi).

b. What is the force acting on the mass?

Formula:

F = Δp / t

Solution:

F = \frac{325}{3}F=

3

325

F = 108.33333...F=108.33333...

F = 100F=100 kg·m/s²

The formula here is derived from Ft = Δp. In order to make the force isolated in the formula, we should divide both sides by time or t.

c. What was the impulse acting on the mass?

Formula:

I = Δp

Solution:

I = 325

I = 300 m/s

Remember, impulse is moving momentum. Therefore, I = Δp is derived from that statement.