<u>Momentum</u>
- a vector quantity; has both magnitude and direction
- has the same direction as object's velocity
- can be represented by components x & y.
Find linebacker momentum given m₁ = 120kg, v₁ = 8.6 m/s north
P₁ = m₁v₁
P₁ = (120)(8.6)
[ P₁ = 1032 kg·m/s ] = y-component, linebacker momentum
Find halfback momentum given m₂ = 75kg, v₂ = 7.4 m/s east
P₂ = m₂v₂
P₂ = (75)(7.4)
[ P₂ = 555 kg·m/s ] = x-component, halfback momentum
Find total momentum using x and y components.
P = √(P₁)² + (P₂)²
P = √(1032)² + (555)²
[[ P = 1171.77 kg·m/s ]] = magnitude
!! Finally, to find the magnitude of velocity, take the divide magnitude of momentum by the total mass of the players.
P = mv
P = (m₁ + m₂)v
1171.77 = (120 + 75)v <em>[solve for v]</em>
<em />v = 1171.77/195
v = 6.0091 ≈ 6.0 m/s
If asked to find direction, take inverse tan of x and y components.
tanθ = (y/x)
θ = tan⁻¹(1032/555)
[ θ = 61.73° north of east. ]
The magnitude of the velocity at which the two players move together immediately after the collision is approximately 6.0 m/s.
Answer:
x ’= 368.61 m, y ’= 258.11 m
Explanation:
To solve this problem we must find the projections of the point on the new vectors of the rotated system θ = 35º
x’= R cos 35
y’= R sin 35
The modulus vector can be found using the Pythagorean theorem
R² = x² + y²
R = 450 m
we calculate
x ’= 450 cos 35
x ’= 368.61 m
y ’= 450 sin 35
y ’= 258.11 m
To do this you want to solve for one variable at a time. So we want to cancel out a variable. Lets cancel x. I will multiply the first equation by the number 4 to get 4y=4x-16.
Now lets solve equation 2 for y, giving
-3y=-4x+3 now add equation 1 to equation 2
Y =-13
Now plug that back in to either
-13=x-4
X=-9
So the answer is (-9,-13)
this question is unclear. please specify.
