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arlik [135]
3 years ago
5

Imagine two billiard balls on a pool table. Ball A has a mass of 2 kilograms and ball B has a mass of 3 kilograms. The initial v

elocity of ball A is 9 meters per second to the right, and the initial velocity of the ball B is 6 meters per second to the left. The final velocity of ball A is 9 meters per second to the left, while the final velocity of ball B is 6 meters per second to the right.
1. Explain what happens to each ball after the collision. Why do you think this occurs? Which of Newton’s laws does this represent?

2. What can you say about the total momentum before and after the collision?

3. What do you think would happen to the velocity of each ball after the collision if the masses and initial velocities of each ball were the same?

4. The mass of ball A is 10 kilograms and the mass of ball B is 5 kilograms. If the initial velocity is set to 3 meters per second for each ball, what is the final velocity of ball B if the final velocity of ball A is 2 meters per second? Use the elastic collision equation to find the final velocity of ball B. Assume ball A initially moves from right to left and ball B moves in the opposite direction. Identify each mass, velocity, and unknown. Show your work, including units, and indicate the direction of ball B in your answer.

5. If the mass of each ball were the same, but the velocity of ball A were twice as much as ball B, what do you think would happen to the final velocity of each ball after the collision? To answer this question, create a hypothesis in the form of an if-then statement. The “if” is the independent variable, or the thing that is being changed. The “then” is the dependent variable, or what you will measure as the outcome.
(Please show your work if you can)
Physics
1 answer:
Orlov [11]3 years ago
3 0
1. After the collision, their velocities have switched. This is Newton's third law of motion.

2. The total momentum is conserved.

3. The same thing would happen if the collision is totally elastic.

For 4 and 5.
Using the conservation of momentum equation
m1v1 + m2v2 = m1'v1' + m2'v2'<span />
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A mass is undergoing simple harmonic motion. When its displacement is 0, it is at its equilibrium position. At that moment, its
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When X = 0

Speed = maximum    V (max) = ω A

Acceleration = zero       a(max) = - ω^2 A

From x = A sin ω t        sin = 0     so displacement = zero

V = ω A cos ω t             cos = 1 and speed = maximum

a = - A ω^2 sin ω t            sin = 0 and acceleration = zero

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2 years ago
A 2.45-kg frictionless block is attached to an ideal spring with force constant 355 N/m. Initially the spring is neither stretch
ANTONII [103]

Answer:

A.    A = 0.913 m

B.    amax = 132.24m/s^2

C.    Fmax = 324.01N

Explanation:

When the block is moving at the equilibrium point , its velocity is maximum.

A. To find the amplitude of the motion you use the following formula for the maximum velocity:

v_{max}=A\omega          (1)

vmax = maximum velocity = 11.0 m/s

A: amplitude of the motion = ?

w: angular frequency = ?

Then, you have to calculate the angular frequency of the motion, by using the following formula:

\omega=\sqrt{\frac{k}{m}}           (2)

k: spring constant = 355 N/m

m: mass of the object = 2.54 kg

\omega = \sqrt{\frac{355N/m}{2.45kg}}=12.03\frac{rad}{s}

Next, you solve the equation (1) for A and replace the values of vmax and w:

A=\frac{v}{\omega}=\frac{11.0m/s}{12.03rad/s}=0.913m

The amplitude of the motion is 0.913m

B. The maximum acceleration of the block is given by:

a_{max}=A\omega^2 = (0.913m)(12.03rad/s)^2=132.24\frac{m}{s^2}

The maximum acceleration is 132.24 m/s^2

C. The maximum force is calculated by using the second Newton law and the maximum acceleration:

F_{max}=ma_{max}=(2.45kg)(132.24m/s^2)=324.01N

It is also possible to calculate the maximum force by using:

Fmax = k*A = (355N/m)(0.913m) = 324.01N

The maximum force exertedbu the spring on the object is 324.01 N

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3 years ago
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Answer:

It will remain same i.e. 4.0 Hz

Explanation:

In a mass-and-spring system, the frequency depends upon the mass (m) and the spring constant (k).

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

Since, the frequency does not depend on the initial displacement of the mass, the frequency would remain the same i.e. 4.0 Hz.

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Answer:

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here y = 3.4 , x = 5 , θ = 56

equation of trajectory

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= 9.27 m

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= 9.27 - 5

= 4.27 m

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