Answer:
total momentum = 8.42 kgm/s
velocity of the first cart is 3.660 m/s
Explanation:
Given data
mass m1 = 2.3 kg
mass m2 = 1.5 kg
final velocity V2 = 4.9 m/s
final velocity V3 = - 1.9 m/s
to find out
total momentum and velocity of the first cart
solution
we know mass and final velocty
and initial velocity of second cart V1 = 0
so now we can calculate total momentum that is m1 v2 + m2 v2
total momentum = 2.3 ×4.9 + 1.5 ×(-1.9)
total momentum = 8.42 kgm/s
and
conservation of momentum is
m1 V + m2 v1 = m1 v2 + m2 v3
put all value and find V
2.3 V + 1.5 ( 0) = 2.3 ( 4.9 ) + 1.5 ( -1.9)
V = 8.42 / 2.3
V = 3.660 m/s
so velocity of the first cart is 3.660 m/s
<span>its reflection of light from a glass surface
</span>
The momentum of the red cart before the collision is 0.2 kgm/s and the blue cart is 0.
The momentum of the red cart after the collision is 0.05 kgm/s and the blue cart is 0.15 kgm/s.
The change in momentum of the system of the carts is 0.
<h3>
Initial momentum of the carts before collision</h3>
The momentum of the carts before the collision is calculated as follows;
P(red) = 0.5 kg x 0.4 m/s = 0.2 kgm/s
P(blue) = 1.5 x 0 = 0
<h3>Momentum of the carts after collision</h3>
The momentum of the carts after the collision is calculated as follows;
P(red) = 0.5 x 0.1 = 0.05 kgm/s
P(blue) = 1.5 0.1 = 0.15 kgm/s
<h3>Change in momentum of the carts</h3>
ΔP = (0.05 + 0.15) - (0.2)
ΔP = 0
Learn more about momentum here: brainly.com/question/7538238
Answer:
F = 147,78*10⁻⁹ [N]
Explanation:
By symmetry the Fy components of the forces acting on charge in point x = 0,7 m canceled each other, and the total force will be twice Fx ( Fx is x axis component of one of the forces .
The angle β ( angle between the line running through one of the charges in y axis and the charge in x axis) is
tan β = 0,5/0,7
tan β = 0,7142 then β = arctan 0,7142 ⇒ β = 35 ⁰
cos β = 0,81
d = √ (0,5)² + (0,7)² d1stance between charges
d = √0,25 + 0,49
d = √0,74 m
d = 0,86 m
Now Foce between two charges is:
F = K* q₁*q₂/ d² (1)
Where K = 9*10⁹ N*m²/C²
q₁ = 2,5* 10⁻⁹C
q₂ = 3,0*10⁻⁹C
d² = 0,74 m²
Plugging these values in (1)
F = 9*10⁹* 2,5* 10⁻⁹*3,0*10⁻⁹ / 0,74 [N*m²/C²]*C*C/m²
F = 91,21 * 10⁻⁹ [N]
And Fx = F*cos β
Fx = 91,21 * 10⁻⁹ *0,81
Fx =73,89*10⁻⁹ [N]
Then total force acting on charge located at x = 0,7 m is:
F = 2* Fx
F = 2*73,89*10⁻⁹ [N]
F = 147,78*10⁻⁹ [N]
