Answer:
Taking the x axis to the right and the y axis to be up, the total change of momentum is [tex]\Delta \vec{ p} = 3.9429 \frac{kg \ m}{s} \hat{j} [/tex]
Explanation:
The momentum [tex]\vec{p}[/tex] is given by:
[tex]\vec{p} = m \ \vec{v}[/tex]
where m is the mass and [tex]\vec{v}[/tex] is the velocity. Now, taking the suffix i for the initial condition, and the suffix f for the final condition, the change in momentum will be:
[tex]\Delta \vec{ p} = \vec{p}_f - \vec{p}_i[/tex]
[tex]\Delta \vec{ p} = m \ \vec{v}_f - m \ \vec{v}_i[/tex]
[tex]\Delta \vec{ p} = m (\ \vec{v}_f - \ \vec{v}_i)[/tex]
As we know the mass of the ball, we just need to find the initial and final velocity.
Knowing the magnitude and direction of a vector, we can obtain the Cartesian components with the formula
[tex]\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )[/tex]
where [tex]| \vec{A} |[/tex] is the magnitude of the vector and θ is the angle measured from the x axis.
Taking the x axis to the right and the y axis to be up, the initial velocity will be:
[tex]\vec{v}_i = 4 \frac{m}{s} ( \ cos ( - (90 \ °- 31 \°)) , sin( - (90 \ ° - 31\°) ) ) =[/tex]
where minus sign appears cause the ball is going downward, and we subtracted the 31 ° as it was measured from the y axis
So, the initial velocity is
[tex]\vec{v}_i = 4 \frac{m}{s} ( \ cos ( - 59 \°) , sin( - 59 \°)) =[/tex]
[tex]\vec{v}_i = ( \ 2.0601 \ \frac{m}{s} , - 3.4286 \frac{m}{s}) =[/tex]
The final velocity is
[tex]\vec{v}_i = 4 \frac{m}{s} ( \ cos ( 90 \ °- 31 \°) , sin( 90 \ ° - 31\°)) =[/tex]
[tex]\vec{v}_i = 4 \frac{m}{s} ( \ cos ( 59 \°) , sin( 59 \°)) =[/tex]
[tex]\vec{v}_i = ( \ 2.0601 \ \frac{m}{s} , 3.4286 \frac{m}{s}) =[/tex]
So, the change in momentum will be
[tex]\Delta \vec{ p} = m (\ \vec{v}_f - \ \vec{v}_i)[/tex]
[tex]\Delta \vec{ p} = 0.575 \ kg (\ ( \ 2.0601 \ \frac{m}{s} , 3.4286 \frac{m}{s} - ( \ 2.0601 \ \frac{m}{s} , - 3.4286 \frac{m}{s}))[/tex]
[tex]\Delta \vec{ p} = 0.575 \ kg (\ ( \ 2.0601 \ \frac{m}{s} - \ 2.0601 \ \frac{m}{s}, 3.4286 \frac{m}{s} + 3.4286 \frac{m}{s}) [/tex]
[tex]\Delta \vec{ p} = 0.575 \ kg (\ 0 , 2 * 3.4286 \frac{m}{s} ) [/tex]
[tex]\Delta \vec{ p} = 0.575 \ kg * 2 * 3.4286 \frac{m}{s} \hat{j} [/tex]
[tex]\Delta \vec{ p} = 3.9429 \frac{kg \ m}{s} \hat{j} [/tex]
