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3 years ago

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What is the mass of a large ship that has a momentum of 1.60×109kg·m/s, when the ship is moving at a speed of 48.0 km/h? (b) Com

pare the ship’s momentum to the momentum of a 1100-kg artillery shell fired at a speed of 1200 m/s.

1 answer:

erastova [34]3 years ago

6 0

a) The mass of the ship is $1.2\cdot 10^8 kg$

b) The ship has a larger momentum than the shell

Explanation:

a)

The momentum of an object is given by:

$p=mv$

where

m is the mass of the object

v is its velocity

For the ship in this problem, we have

$p=1.60\cdot 10^9 kg m/s$ is the momentum

$v=48.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=13.3 m/s$ is the velocity

Solving for m, we find the mass of the ship:

$m=\frac{p}{v}=\frac{1.60\cdot 10^9}{13.3}=1.2\cdot 10^8 kg$

b)

The momentum of the artillery shell is given by

$p=mv$

where

m is its mass

v is its velocity

For the shell in this problem,

m = 1100 kg

v = 1200 m/s

Substituting,

$p=(1100)(1200)=1.32\cdot 10^6 kg m/s$

So, we see that the ship has a larger momentum.

Learn more about momentum:

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∴

inputting the values into the equation above, we have;

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$\mathsf{0.3 =4.9*(t^2)}$

By dividing both sides by 4.9, we have:

$\mathsf{t^2 = \dfrac{0.3}{4.9}}$

$\mathsf{t^2 = 0.0612}$

$\mathsf{t = \sqrt{0.0612}}$

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However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

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