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3 years ago

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A tennis ball (m=0.060 kg) is moving horizontally at 20 m/s toward a tennis player who hits it straight back at 26 m/s. What is

the change in momentum (in kg m/s) delivered to the tennis ball? 20 m/s 26 m/s
A. 0.36
B. 12.2
C. 1.84
D. 2.76

1 answer:

Rus_ich [418]3 years ago

4 0

Answer:

0.36 kg-m/s

Explanation:

Given that,

Mass of a ball, m = 0.06 kg

Initial velocity of the ball, u = 20 m/s

Final velocity of the ball, v = 26 m/s

We need to find the change in momentum of the tennis ball. It is equal to the final momentum minus initial momentum

$\Delta p=m(v-u)\\\\=0.06\times (26-20)\\\\=0.36\ kg-m/s$

So, the change in momentum of the ball is 0.36 kg-m/s.

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Answer:

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Explanation:

We are given that

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According to question

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$Tr=I\alpha$

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$T=\frac{1}{2}m_1a=\frac{1}{2}(3.3)a$

Substitute the value

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$48.02=a(4.9+1.65)=6.55a$

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A) What is the work done?

Answer: We need to use the formula

$w=-F_f(d)$

$w=-(35)(20)$

$w=-700J$

B) What is the work done on the cart by the gravitational force?

Alright, we know that the gravitional force is perpendicular to the diplacement. Therefore, we gonna use the following formula:

$w=Fdcos90$

$w=0$

C) What is the work done on the cart by the shopper?

This is the easier part, since we already know that the work done by the shopper is the same as the work done by the friction force

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D) Find the force the shopper exerts, using energy considerations.

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E) What is the total work done?

You just need to add them:

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Answer:

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