Answer:
a) v₁fin = 3.7059 m/s (→)
b) v₂fin = 1.0588 m/s (→)
Explanation:
a) Given
m₁ = 0.5 Kg
L = 70 cm = 0.7 m
v₁in = 0 m/s ⇒ Kin = 0 J
v₁fin = ?
h<em>in </em>= L = 0.7 m
h<em>fin </em>= 0 m ⇒ U<em>fin</em> = 0 J
The speed of the ball before the collision can be obtained as follows
Einitial = Efinal
⇒ Kin + Uin = Kfin + Ufin
⇒ 0 + m*g*h<em>in</em> = 0.5*m*v₁fin² + 0
⇒ v₁fin = √(2*g*h<em>in</em>) = √(2*(9.81 m/s²)*(0.70 m))
⇒ v₁fin = 3.7059 m/s (→)
b) Given
m₁ = 0.5 Kg
m₂ = 3.0 Kg
v₁ = 3.7059 m/s (→)
v₂ = 0 m/s
v₂fin = ?
The speed of the block just after the collision can be obtained using the equation
v₂fin = 2*m₁*v₁ / (m₁ + m₂)
⇒ v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)
⇒ v₂fin = 1.0588 m/s (→)
Answer:
the answer is gravitational force hope this helps u stay safe
Explanation:
The distance for any rectilinear motion at constant acceleration is:
d = v₀t + 0.5at²
where
v₀ is the initial velocity
So, if v₀ = 6v, and it stopped to 0 m/s, then the acceleration is equal to:
a = (0 - 6v)/t = -6v/t
Thus,
d = (6v)(t) + (0.5)(-6v/t)(t²)
d = 6vt - 3t
<span>d = 3t(2v - 1)</span>
<h2>
Answer: Gravitational attraction will be the same</h2>
According to the law of universal gravitation, which is a classical physical law that describes the gravitational interaction between different bodies with mass:
(1)
Where:
is the module of the force exerted between both bodies
is the universal gravitation constant.
and 
are the masses of both bodies.
is the distance between both bodies
Now, if we double both masses and the distance also doubles, this means:
and will be now 
and 
will be now 
Let's rewrite the equation (1) with this new values:
(2)
Solving and simplifying:
(3)
As we can see, equation (3) is the same as equation (1).
So, if the masses both double and the distance also doubles the <u>Gravitational attraction between both masses will remain the same.</u>
