Sayid made a chart listing data of two colliding objects. A 5-column table titled Collision: Two Objects Stick Together with 2 r
2 answers:
Answer:
6 m/s is the missing final velocity
Explanation:
From the data table we extract that there were two objects (X and Y) that underwent an inelastic collision, moving together after the collision as a new object with mass equal the addition of the two original masses, and a new velocity which is the unknown in the problem).
Object X had a mass of 300 kg, while object Y had a mass of 100 kg.
Object's X initial velocity was positive (let's imagine it on a horizontal axis pointing to the right) of 10 m/s. Object Y had a negative velocity (imagine it as pointing to the left on the horizontal axis) of -6 m/s.
We can solve for the unknown, using conservation of momentum in the collision: Initial total momentum = Final total momentum (where momentum is defined as the product of the mass of the object times its velocity.
In numbers, and calling the initial momentum of object X and
the initial momentum of object Y, we can derive the total initial momentum of the system:
Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):
Final momentum of the system:
We then set the equality of the momenta (total initial equals final) and proceed to solve the equation for the unknown(final velocity of the system):
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Details are missing in the question. Complete text of the problem:
"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.
"The gravitational force exerted on a baseball is 2.28 N down. A pitcher throws the ball horizontally with velocity 16.5 m/s by uniformly accelerating it along a straight horizontal line for a time interval of 181 ms. The ball starts from rest.
(a) Through what distance does it move before its release? (m)
(b) What are the magnitude and direction of the force the pitcher exerts on the ball? (Enter your magnitude to at least one decimal place.)"
Solution
(a) The pitcher accelerates the baseball from rest to a final velocity of , so
, in a time interval of
. The acceleration of the ball in the horizontal direction (x-axis) is therefore
And the distance covered by the ball during this time interval, before it is released, is:
(b) For this part we need to consider also the weight of the ball, which is
From this, we find its mass:
Now we can calculate the magnitude of the force the pitcher exerts on the ball. On the x-axis, we have
We also know that the ball is moving straight horizontally. This means that the vertical component of the force exerted by the pitcher must counterbalance the weight of the ball (acting downward), in order to have a net force of zero along the y-axis, and so:
(upward)
So, the magnitude of the force is
To find the direction, we should find the angle of F with respect to the horizontal. This is given by
From which we find
Answer:
Explanation:
As we know that the mass is revolving with constant angular speed in the circle of radius R
So we will have
now the position vector at a given time is
now the linear velocity is given as