A block of wood 3 cm on each
1 answer:
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The motion of the ball on the vertical axis is an accelerated motion, with acceleration
The following relationship holds for an uniformly accelerated motion:
where S is the distance covered, vf the final velocity and vi the initial velocity.
If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
So we can rewrite the equation as
from which we can isolate h
(1)
Now let's assume that
is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: 
. So the maximum height of the second ball is
(2)
Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
The following relationship holds for an uniformly accelerated motion:
where S is the distance covered, vf the final velocity and vi the initial velocity.
If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
So we can rewrite the equation as
from which we can isolate h
Now let's assume that
Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
Answer:
a) v₂ = 4.2 m/s
b) v₂ = 5 m/s
Explanation:
a)
We will use the law of conservation of momentum here:
where,
m₁ = m₂ = mass of bowling pin = 1.8 kg
u₁ = speed of first pin before collsion = 5 m/s
u₂ = speed of second pin before collsion = 0 m/s
v₁ = speed of first pin after collsion = 0.8 m/s
v₂ = speed of second after before collsion = ?
Therefore,
<u>v₂ = 4.2 m/s</u>
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b)
We will use the law of conservation of momentum here:
where,
m₁ = m₂ = mass of bowling pin = 1.8 kg
u₁ = speed of first pin before collsion = 5 m/s
u₂ = speed of second pin before collsion = 0 m/s
v₁ = speed of first pin after collsion = 0 m/s
v₂ = speed of second after before collsion = ?
Therefore,
<u>v₂ = 5 m/s</u>