# Calculate the magnitude of the linear momentum for the following cases:

1. A proton with mass 1.67X10^(-27) kg, moving with a speed 5X10^(6) m/s.
2. A 15.0g bullet moving with a speed of 300m/s.
3. A 75.0 kg sprinter running with a speed of 10.0 m/s.
4. Earth (mass = 5.98X10^(24) kg) moving with an orbital speed equal to 2.98X10^(4) m/s.

The aim of this question is to learn the calculations involved in determining the linear momentum of a moving object.

The linear momentum of an object of mass m kilogram moving with a linear speed of v meters per second is defined as the product of mass  m and the speed v. Mathematically:

$P \ = \ m v$

Part (a): A proton with mass $1.67 \times 10^{ -27 } \ kg$, moving with a speed of $5 \times 10^{ 6 } \ m/s$.

Here:

$m \ = \ 1.67 \times 10^{ -27 } \ kg$

And:
$v \ = \ 5 \times 10^{ 6 } \ m/s$

So:

$P \ = \ m v$

$\Rightarrow P \ = \ ( 1.67 \times 10^{ -27 } \ kg )( 5 \times 10^{ 6 } \ m/s )$

$\Rightarrow P \ = \ 8.35 \times 10^{ -21 } \ kg \ m/s$

Part (b): A $15.0 \ g$ bullet moving with a speed of $300 \ m/s$.

Here:

$m \ = \ 0.015 \ kg$

And:
$v \ = \ 300 \ m/s$

So:

$P \ = \ m v$

$\Rightarrow P \ = \ (0.015 \ kg )( 300 \ m/s )$

$\Rightarrow P \ = \ 4.5 \ kg \ m/s$

Part (c): A $75.0$ $kg$ sprinter running with a speed of $10.0$  $m/s$.

Here:

$m \ = \ 75.0 \ kg$

And:
$v \ = \ 10.0 \ m/s$

So:

$P \ = \ m v$

$\Rightarrow P \ = \ (75.0 \ kg )( 10.0 \ m/s )$

$\Rightarrow P \ = \ 750.0 \ kg \ m/s$

Part (d): Earth $( \ mass \ = \ 5.98 \times 10^{24} \ kg \ )$ moving with an orbital speed equal to $2.98 \times 10^{4} \ m/s$.

Here:

$m \ = \ 5.98 \times 10^{24}\ kg$

And:
$v \ = \ 2.98 \times 10^{4} \ m/s$

So:

$P \ = \ m v$

$\Rightarrow P \ = \ ( 5.98 \times 10^{24} \ kg )( 2.98 \times 10^{4} \ m/s )$

$\Rightarrow P \ = \ 1.78 \times 10^{29} \ kg \ m/s$

## Numerical Result

$\text{Part (a): } P \ = \ 8.35 \times 10^{ -21 } \ kg \ m/s$

$\text{Part (b): } P \ = \ 4.5 \ kg \ m/s$

$\text{Part (c): } P \ = \ 750.0 \ kg \ m/s$

$\text{Part (d): } P \ = \ 1.78 \times 10^{29} \ kg \ m/s$

## Example

Calculate the magnitude of the linear momentum for an object of mass $5 \ kg$ moving with a speed of $80 \ m/s$.

Here:

$m \ = \ 5 \ kg$

And:
$v \ = \ 80 \ m/s$

So:

$P \ = \ m v$

$\Rightarrow P \ = \ (5 \ kg )( 80 \ m/s ) \ = \ 400 \ kg \ m/s$