Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Answer:
V=15.3 m/s
Explanation:
To solve this problem, we have to use the energy conservation theorem:

the elastic potencial energy is given by:

The work is defined as:

this work is negative because is opposite to the movement.
The gravitational potencial energy at 2.5 m aboves is given by:

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

Decrease because its 0.0 m/s
Answer
given,
mass of ball, m = 57.5 g = 0.0575 kg
velocity of ball northward,v = 26.7 m/s
mass of racket, M = 331 g = 0.331 Kg
velocity of the ball after collision,v' = 29.5 m/s
a) momentum of ball before collision
P₁ = m v
P₁ = 0.0575 x 26.7
P₁ = 1.535 kg.m/s
b) momentum of ball after collision
P₂ = m v'
P₂ = 0.0575 x (-29.5)
P₂ = -1.696 kg.m/s
c) change in momentum
Δ P = P₂ - P₁
Δ P = -1.696 -1.535
Δ P = -3.231 kg.m/s
d) using conservation of momentum
initial speed of racket = 0 m/s
M u + m v = Mu' + m v
M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)
0.331 u' = 3.232
u' = 9.76 m/s
change in velocity of the racket is equal to 9.76 m/s
The force of the racket affects the ball's motion because it changes the momentum of the ball.
<h3>Impulse received by the ball</h3>
The impulse received by the ball through the racket affects the motion because it changes the momentum of the ball.
The ball which is initially at rest, will gain momentum after been hit with the racket.
J = ΔP = Ft
where;
- J is the impulse received by the ball
- ΔP is change in momentum of the ball
- F is the applied force
- t is the time of action
Thus, the force of the racket affects the ball's motion because it changes the momentum of the ball.
Learn more about impulse here: brainly.com/question/25700778