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8090 [49]
3 years ago
14

After polishing his 2-kg wrestling trophy, Mike sets it down on the ground and walks away to find more polish. Meanwhile, Julie

comes around the corner and throws an 8-kg bowling ball towards the unguarded trophy at a velocity of 20 m/s. Before the collision:
What is the initial momentum of the trophy?



What is the initial momentum of the bowling ball?

What is the total momentum of the system before the collision?





What is the total momentum of the system after the collision?




After the collision, the bowling ball slows down to a velocity of 12 m/s. What is the velocity of the trophy?
​
Physics
1 answer:
klio [65]3 years ago
4 0

1) The initial momentum of the trophy is zero

2) The initial momentum of the bowling ball is 160 kg m/s

3) The total momentum before the collision is 160 kg m/s

4) The total momentum of the system after the collision is 160 kg m/s

5) The final velocity of the trophy is 32 m/s

Explanation:

1)

The momentum of an object is given by

p=mv

where

m is the mass of the object

v is its velocity

In this problem, the data for the trophy before the collision are:

m = 2 kg is the mass

v = 0 is its initial velocity

Therefore, the initial momentum of the trophy is

p_1=(2)(0)=0

2)

Using the same equation used in part 1), the initial momentum of the bowling ball is

p=mv

where

m is the mass of the bowling ball

v is its initial velocity

The data of the problem are

m = 8 kg is the mass

v = 20 m/s is the velocity

Substituting,

p_2=(8)(20)=160 kg m/s

3)

The total momentum of the system before the collision is given by the sum between the initial momentum of the trophy and the initial momentum of the bowling ball:

p_i = p_1 + p_2

where

p_1 is the initial momentum of the trophy

p_2 is the initial momentum of the ball

Here we have

p_1 = 0

p_2 = 160 kg m/s

Therefore, the total momentum is

p_i = 0 + 160 = 160 kg m/s

4)

According to the law of conservation of momentum, for an isolated system (=no external unbalanced forces acting on the system), the total momentum of the system is conserved before and after the collision:

p_i = p_f

where

p_i is the total momentum before the collision

p_f is the total momentum after the collision

If we consider the system in the problem to be isolated (i.e. no frictional forces acting on the ball or the trophy), we can therefore say that the total momentum after the collision must be equal to the total momentum before the collision: therefore,

p_f = 160 kg m/s

5)

We can write the total momentum after the collision as

p_f = m_1 v_1 + m_2 v_2

where:

m_1 = 2 kg is the mass of the trophy

v_1 is the final velocity of the trophy

m_2 = 8 kg is the mass of the bowling ball

v_2 = 12 m/s is the final velocity of the ball

Since we also know the value of the final total momentum,

p_f = 160 kg m/s

we can solve the equation to find the velocity of the trophy:

v_1 = \frac{p_f - m_2 v_2 }{m_1}=\frac{160-(8)(12)}{2}=32 m/s

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

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a).

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F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 381.44 N

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6 0
3 years ago
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Answer:

Explanation:

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7 0
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Answer:

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Explanation:

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The speed at which the helicopter ascends is not specified in the statement, but according to a similar problem, we will use 12.5 \frac{m}{s}.

First, we must calculate the time and the maximum height at which the camera arrives after leaving the helicopter.

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Where:

V_{f}: final speed at maximum height.

V_{0}: initial speed when it falls from the helicopter.

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h: height reached from 60 m when leaving the helicopter

as V_{f}=0

0=(12,5\frac{m}{s}) ^{2} - 2 * 9,8\frac{m}{s^{2} } * h

clear h:

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0=12.5 \frac{m}{s} - 9.8 \frac{m}{s^{2} } * t

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t=12.5 \frac{m}{s} / 9.8 \frac{m}{s^{2} }

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The equation we will use is Y=v_{0}*t+(\frac{g*t^{2} }{2} )

where:

t: time it takes for the camera to fall.

Y: height from where the camera falls concerning the ground.

v_{0}: initial speed of the camera at the time of starting the fall.

g: acceleration of gravity, estimated at 9.8 \frac{m}{s^{2} }

Step 1: As the helicopter ascends with constant speed, the initial speed of the camera at the moment of falling is 0.

v_{0}=0

So the first term of our equation is nullified.

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Step 3: I replace the values with the incognites and get "t".

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The total time it takes for the camera to fall from the moment it leaves the helicopter is the sum of the time it takes to reach the maximum point of height and the time it takes to fall to the ground from that height.

t= 1,27 s + 3,73 s = 5 s

Have a nice day!

3 0
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1.

B. The person and the Earth

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