Answer:
The rise in height of combined block/bullet from its original position is 0.45m
Explanation:
Given;
mass of bullet, m₁ = 12 g = 0.012 kg
mass of block of wood, m₂ = 1 kg
initial speed of bullet, u₁ = 250 m/s.
initial speed of block of wood, u₂ = 0
From the principle of conservation of linear momentum, calculate the final speed of the combined block/bullet system.
m₁u₁ + m₂u₂ = v(m₁+m₂)
where;
v is the final speed of the combined block/bullet system.
0.012 x 250 + 0 = v (0.012 + 1)
3 = v (1.012)
v = 3/1.012
v = 2.96 m/s
From the principle of conservation of energy, calculate the rise in height of the block/bullet combined from its original position.
¹/₂mv² = mgh
¹/₂v² = gh
¹/₂ (2.96)² = (9.8)h
4.3808 = 9.8h
h = 4.3808/9.8
h = 0.45 m
Therefore, the rise in height of combined block/bullet from its original position is 0.45m
Given:
(Initial velocity)u=20 m/s
At the maximum height the final velocity of the ball is 0.
Also since it is a free falling object the acceleration acting on the ball is due to gravity g.
Thus a=- 9.8 m/s^2
Now consider the equation
v^2-u^2= 2as
Where v is the final velocity which is measured in m/s
Where u is the initial velocity which is measured in m/s
a is the acceleration due to gravity measured in m/s^2
s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.
Substituting the given values in the above formula we get
0-(20x20)= 2 x- 9.8 x s
s= 400/19.6= 20.41m
Thus the maximum height attained is 20.41 m by the ball
If the moon was hit by an asteroid there would be a crater mark and possible movement.
Answer:
Explanation:
If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.
mgh = ½mv²
v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s
However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy
mgh = ½mv² + ½Iω²
mgh = ½mv² + ½(½mR²)(v/R)²
2gh = v² + ½v²
2gh = 3v²/2
v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s
