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yKpoI14uk [10]
3 years ago
12

Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. The initial v

elocity of the ball A is 6 meters per second to the right, and the initial velocity of the ball B is 12 meters per second to the left.
1. When the two balls hit each other, what will happen if it is a perfectly
inelastic collision?

2. Compare the final velocity of the balls.

3. What can you say about the total momentum before and after the collision?
Physics
1 answer:
wlad13 [49]3 years ago
3 0
1) In a perfectly inelastic collision, the two balls stick together after the collision. In this type of collision, the total kinetic energy of the system is not conserved, while the total momentum is conserved.
If we callv_f the final velocity of the two balls that stick together, the conservation of the total momentum before and after the collision can be written as
m_a v_{Ai} + m_b v_{Bi} = (m_A+m_B)v_f (1)
where
m_A=7 kg is the mass of ball A
m_B=2 kg is the mass of ball B
v_{Ai}=6 m/s is the initial velocity of ball A
v_{Bi}=-12 m/s is the initial velocity of ball B (taken with a negative sign, since it goes in the opposite direction of ball A)

If we solve (1) to find v_f, we find that the final velocity of the balls is
v_f= \frac{m_Av_{Ai}+m_Bv_{Bi}}{m_A+m_B}= \frac{(7\cdot 6)+(2 \cdot (- 12))}{7+2}= \frac{18}{9}=2 m/s
and the positive sign means the two balls are going to the right.


2) I assume here we are talking about an elastic collision. In this case, both total momentum and total kinetic energy are conserved:
m_A v_{Ai}+m_B v_{Bi} = m_A v_{fA} + m_B v_{fB}
\frac{1}{2}m_A v_{Ai}^2+ \frac{1}{2}m_B v_{Bi}^2= \frac{1}{2}m_Av_{fA}^2+ \frac{1}{2}m_B v_{fB}^2
where
v_{fA} is the final velocity of ball A
v_{fB} is the final velocity of ball B

If we solve simultaneously the two equations, we find:
v_{fA}= \frac{v_{Ai}(m_A-m_B)+2m_Bv_{Bi}}{m_A+m_B} = \frac{(6)(7-2)+2(2)(-12)}{7+2}=-2 m/s
v_{fB}= \frac{v_{Bi}(m_B-m_A)+2m_Av_{Ai}}{m_A+m_B} = \frac{(-12)(2-7)+2(7)(6)}{7+2}= \frac{144}{9}=16 m/s
So, after the collision, ball A moves to the left with velocity v=-2 m/s and ball B moves to the right with velocity v=16 m/s.

3) The total momentum before and after the collision is conserved.
In fact, the total momentum before the collision is:
p_i = m_A v_{A} + m_B v_{fB} = (7\cdot 6)+(2 \cdot (-12))=42-24=18 m/s
and the total momentum after the collision is:
p_f = m_A v_{A} + m_B v_{fB} = (7\cdot (-2))+(2 \cdot 16)=-14+32=18 m/s

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Answer:

13800 N

Explanation:

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F=\frac {I}{t}

Substituting I with 13.8 N.s and time, t witg 0.001 s then the average force is calculated as

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A hot–air balloon is moving at a speed of 10.0 meters/second in the +x–direction. The balloonist throws a brass ball in the +x–d
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Answer:

Option (D) is correct.

Explanation:

The balloon lands horizontally at a distance of 420 m from a point where it as released.

Velocity of air balloon along +X axis =10 m/s

velocity of ball=4 m/s along + X axis

the velocity of balloon gets added to the velocity of ball. So the resultant velocity of the balloon=10+4 = 14 m/s

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The distance traveled is given by d= v t

d= 14 (30)

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Thus the balloon lands horizontally at a distance of 420 m from a point where it as released.

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If kinetic energy of a body is increased by 125%, the percentage increase in Momentum is?
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Why is it a good idea to extend your bare hand forward when you are getting ready to catch a fast-moving baseball?
qaws [65]

Answer:d

Explanation:

It is good to extend your bare hand forward when you are getting ready to catch a fast-moving ball so you can pull your hand back, increasing the time to slow the ball thus decreasing the force.

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Serggg [28]

Answer:

<em>n =1.33 revolutions</em>

Explanation:

<u>Uniform Circular Motion</u>

The angular speed can be calculated in two different ways:

\displaystyle \omega=\frac{v}{r}

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v = tangential speed

r = radius of the circle described by the rotating object

Also:

\omega=2\pi f

Where:

f = frequency

Solving for f:

\displaystyle f=\frac{\omega}{2\pi}

Since the frequency is calculated when the number of revolutions n and the time t are known:

\displaystyle f=\frac{n}{t}

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The particle moves in a circle of r=90 m with a speed v=25 m/s. Thus the angular speed is:

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Now we calculate f:

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