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Licemer1 [7]
2 years ago
12

An object of mass 2.5 kg has a momentum < 3, 6, 7 > kg m/s. At this instant the object is acted on a by a force < 50, 5

0, 100 > N for 5 x 10-3 s . What is the momentum of the object at the end of this time interval?
Physics
1 answer:
rosijanka [135]2 years ago
5 0

Answer:

The momentum of the object at the end is (3.25i+6.25j+7.5k)\ kg-m/s

Explanation:

Given that,

Mass of object = 2.5 kg

Momentum p= 3i+6j+7k

Force F=50i+50j+100k

Time t=5\times10^{-3}\ s

We need to calculate the momentum of the object at the end

Using formula of impulse

J=\Delta p

J=m\Delta v...(I)

J=F\times\Delta t....(II)

From equation (I) and (II)

F\times\Delta t=m\Delta v

F\times \Delta t=m(v_{f}-v_{i})

mv_{f}=F\times \Delta t+mv_{i}

Put the value into the formula

p_{f}=(50i+50j+100k)\times5\times10^{-3}+3i+6j+7k

p_{f}=0.25i+0.25j+0.5k+3i+6j+7k

p_{f}=(3.25i+6.25j+7.5k)\ kg m/s

Hence, The momentum of the object at the end is (3.25i+6.25j+7.5k)\ kg-m/s

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Answer:

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Thickness of the tennis ball, t = 2.0 mm = 2.0\times 10^{- 3}\ m

Max velocity of the tennis ball, v_{m} = 200\ mph = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

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Kinetic energy of the tennis ball, KE' = \frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s

Now, the distance the ball can penetrate to is given by:

\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}

\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js

Thus

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

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P(t) = e^{\frac{- 2t}{\eta}}

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P(t) = e^{-2.63\times 10^{32}}

Taking log on both the sides:

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