<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:
The new height the ball will reach = (1/4) of the initial height it reached.
Explanation:
The energy stored in any spring material is given as (1/2)kx²
This energy is converted to potential energy, mgH, of the ball at its maximum height.
If the initial height reached is H
And the initial compression of the spring = x
So, mgH = (1/2)kx²
H = kx²/2mg
The new compression, x₁ = x/2
New energy of loaded spring = (1/2)kx₁²
And the new potential energy = mgH₁
mgH₁ = (1/2)kx₁²
But x₁ = x/2
mgH₁ = (1/2)k(x/2)² = kx²/8
H₁ = kx²/8mg = H/4 (provided all the other parameters stay constant)
The shuttles acceleration in the creases as the fuel is burned because the acceleration of the obect as produced by net force is directly proportional to the magnitude of the net force.
If the pressure above a solution containing a gas solute is reduced, the limit of the gas's solubility will decrease.
I think you can just sub the values in? unless the qn is asking for smth else?
