Question

Calculate the magnitude of the linear momentum for each of the following cases

Answer 1

Answer:

(a)$p = 1.002x10^{-20}Kg.m/s$

(b)$p = 0.598Kg.m/s$

(c)$p = 94.4Kg.m/s$

(d)$p = 1.77x10^{29}Kg.m/s$

Explanation:

The linear momentum is defined as:

$p = mv$  (1)

Where m is the mass and v is the velocity

a.) A proton with mass $1.67 x10^{-27} kg$ moving with a velocity of $6 x 10^{6} m/s$.

Replacing those values in equation (1) it is gotten:

$p = (1.67x10^{-27}Kg)(6x10^{6}m/s)$

$p = 1.002x10^{-20}Kg.m/s$

So, it has a linear momentum of $1.002x10^{-20}Kg.m/s$

b.) A 1.6 g bullet moving with a speed of 374m/s to the right.

Notice that in this case it is necessary to express the mass of the bullet in terms of kilograms:

$1.6g . \frac{1Kg}{1000g}$ ⇒ $1.6x10^{-3}Kg$

$m = 1.6x10^{-3}Kg$

$p = (1.6x10^{-3}Kg)(374m/s)$

$p = 0.598Kg.m/s$

c.) A 8 kg sprinter running with a velocity of 11.8 m/s.

$p = (8Kg)(11.8m/s)$

$p = 94.4Kg.m/s$

d.) Earth ($m=5.98x10^{24} kg$) moving with an orbital speed equal to 29700 m/s.

$p = (5.98x10^{24}Kg)(29700m/s)$

$p = 1.77x10^{29}Kg.m/s$