In the game of billiards, all the balls have approximately the same mass, about 0.17 kg. In the figure, the cue ball strikes ano
ther ball such that it follows the path shown. The other ball has a speed of 1.5 m/s immediately after the collision. What is the speed (in m/s) of the cue ball after the collision?
1 answer:
Answer:
v = 2.6 m/s
Explanation:
The question is incomplete because the diagram is not given in the question.
The diagram of this question is attached below.
According to the law of conversation of momentum.
mv(initial) = mv(final)
We can see in the diagram that:
mv(initial) = 3 × 0.17
mv(final) = mv₁(final) + mv₂(final)
Consider only the horizontal components of velocities, to compare them to initial velocity.
mv(final) = mv₁(final) + mv₂(final)
0.17 × 3 = (0.17)(v)(cos 30) + (0.17)(1.5)(cos 60)
0.51 = 0.1472v + 0.1275
0.3825 = 0.1472v
v = 2.6 m/s
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Answer:
a)
b)
c)
d) or 18.3 cm
Explanation:
For this case we have the following system with the forces on the figure attached.
We know that the spring compresses a total distance of x=0.10 m
Part a
The gravitational force is defined as mg so on this case the work donde by the gravity is:
Part b
For this case first we can convert the spring constant to N/m like this:
And the work donde by the spring on this case is given by:
Part c
We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:
And if we solve for the initial velocity we got:
Part d
Let d1 represent the new maximum distance, in order to find it we know that :
And replacing we got:
And we can put the terms like this:
If we multiply all the equation by 2 we got:
Now we can replace the values and we got:
And solving the quadratic equation we got that the solution for or 18.3 cm because the negative solution not make sense.
