A 90kg man is standing still on frictionless ice. His friend tosses him a 10kg ball, which has a horizontal velocity of 20m/s. A
2 answers:
Answer:
The man's velocity is
Explanation:
If we want to solve this exercise, we need to use the Momentum Conservation Principle.
The momentum is a vectorial magnitude define as :
Where ''p'' is the momentum
Where ''m'' is the mass of the object
And where ''v'' is the velocity vector.
For any collision occurring in an isolated system, momentum is conserved. We define an isolated system as a system in which the sumatory of exterior forces is equal to zero.
For this exercise, we can solve it as a collision ⇒
Where ''m1'' is the mass of the ball
Where ''v1'' is the velocity vector of the ball
Where ''m2'' is the mass of the man
And where ''v2'' is the velocity vector of the man after catching the ball.
If we replace all the data in the equation :
This will be the magnitude of the velocity vector from the man.
If we define as positive the sense in which his friend tosses him the ball (given that we know the direction), the man's velocity will be
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Answer:
<em>The end of the ramp is 38.416 m high</em>
Explanation:
<u>Horizontal Motion </u>
When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.
The maximum horizontal distance traveled by the object can be calculated as follows:
If the maximum horizontal distance is known, we can solve the above equation for h:
The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:
h= 38.416 m
The end of the ramp is 38.416 m high
A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force
, directed toward the center of the circle (so, horizontally)
- the weight of the plane:
, downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)
The radius of the circle is
, so the centripetal force acting on the plane is
On the vertical axis, we have two forces: the weight
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to
, we can find the magnitude of this other force F by using Newton's second law:
So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
And so, the angle is
- the centripetal force
- the weight of the plane:
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)
The radius of the circle is
On the vertical axis, we have two forces: the weight
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to
So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
And so, the angle is