Physics, 14.06.2021 20:20 gungamer720

A bullet of mass 0.5 kg is moving horizontally with a speed of 50 m/s when it hits a block of mass 3 kg that is at rest on a horizontal surface with a coefficient of friction of 0.2.
After the collision the bullet becomes embedded in the block.
A) What is the net momentum of the bullet-block system before the collision?
B) Find the total energy of the bullet-block system before the collision?
C) What is the speed of the bullet-block system after the collision?
D) Find the total energy of the bullet-block system after the collision?
E) How much work must be done to stop the bullet-block system?
F) *Find the maximum traveled distance of the bullet-block after the collision?

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Answer from: jazzy9438

a) The net momentum of the bullet-block system before the collision is 25 kilogram-meters per second.

b) The initial translational kinetic energy of the bullet before the collision is 625 joules.

c) The final speed of the bullet-block system after the collision is 7.143 meters per second.

d) The total energy of the bullet-block system after the collision is 89.289 joules.

e) 89.289 joules must be done to stop the bullet-block system.

f) The bullet-block system will travel 13.007 meters before stopping.

Explanation:

a) Since no external forces are applied on the system defined by the bullet and the block, then the net momentum is conserved and can be calculated by the initial momentum of the bullet:

$p = m\cdot v_{o}$ (1)

Where:

- Net momentum, in kilogram-meters per second.

- Mass of the bullet, in kilograms.

$v_{o}$ - Initial speed of the bullet, in meters per second.

If we know that $m = 0.5\,kg$ and $v_{o} = 50\,\frac{m}{s}$ , then the net momentum of the bullet-block system before the collision is:

$p = (0.5\,kg)\cdot \left(50\,\frac{m}{s} \right)$

$p = 25\,\frac{kg\cdot m}{s}$

The net momentum of the bullet-block system before the collision is 25 kilogram-meters per second.

b) The total energy of the bullet before the collision is its initial translational kinetic energy (), in joules:

$K = \frac{1}{2}\cdot m \cdot v_{o}^{2}$ (2)

$K = \frac{1}{2}\cdot (0.5\,kg)\cdot \left(50\,\frac{m}{s} \right)^{2}$

$K = 625\,J$

The initial translational kinetic energy of the bullet before the collision is 625 joules.

c) Both the bullet and the block experiments a complete inelastic collision, then the final speed of the bullet-block system is calculated solely by the Principle of Momentum Conservation:

$v_{f} = \frac{m\cdot v_{o}}{m+M}$ (3)

Where:

$v_{f}$ - Final speed, in meters per second.

- Mass of the block, in kilograms.

If we know that $m = 0.5\,kg$ , $v_{o} = 50\,\frac{m}{s}$ and $M = 3\,kg$ , then the final speed of the bullet-block system is:

$v_{f} = \left(\frac{0.5\,kg}{0.5\,kg + 3\,kg} \right)\cdot \left(50\,\frac{m}{s} \right)$

$v_{f} = 7.143\,\frac{m}{s}$

The final speed of the bullet-block system after the collision is 7.143 meters per second.

d) The total energy of the bullet-block system after the collision is the translational kinetic energy of the system (), in joules, is:

$K = \frac{1}{2}\cdot (m + M)\cdot v_{f}^{2}$ (4)

$K = \frac{1}{2}\cdot (0.5\,kg + 3\,kg)\cdot \left(7.143\,\frac{m}{s} \right)^{2}$

$K = 89.289\,J$

The total energy of the bullet-block system after the collision is 89.289 joules.

e) By Work-Energy Theorem, magnitude of the work done by friction must be equal to the magnitude of the translational kinetic energy of the system. Hence, 89.289 joules must be done to stop the bullet-block system.

f) The maximum travelled distance of the bullet-block after the collision can be determined by means of Work-Energy Theorem and definition of work:

$W = \mu_{k}\cdot (m+M)\cdot g\cdot s$ (5)

Where:

- Work done by friction, in joules.

- Gravitational acceleration, in meters per square second.

- Travelled distance, in meters.

$\mu_{k}$ - Kinetic coefficient of friction, no unit.

If we know that $m = 0.5\,kg$ , $M = 3\,kg$ , $\mu_{k} = 0.2$ , $g = 9.807\,\frac{m}{s^{2}}$ and $W = 89.289\,J$ , then the travelled distance of the bullet-block system is:

$s = \frac{W}{\mu_{k}\cdot (m+M)\cdot g}$

$s = \frac{89.289\,J}{0.2\cdot (0.5\,kg + 3\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$s = 13.007\,m$

The bullet-block system will travel 13.007 meters before stopping.

Answer from: Quest

Ihave no idea sorry

Answer from: Quest

Neptune is the eighth planet from the sun. it was the first planet to get its existence predicted by mathematical calculations before it was actually seen through a telescope on sept. 23, 1846. irregularities in the orbit of uranus led french astronomer alexis bouvard to suggest that the gravitational pull from another celestial body might be responsible. german astronomer johann galle then relied on subsequent calculations to spot neptune via telescope. previously, astronomer galileo galilei sketched the planet, but he mistook it for a star due to its slow motion. in accordance with all the other planets seen in the sky, this new world was given a name from greek and roman mythology — neptune, the roman god of the sea.

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A bullet of mass 0.5 kg is moving horizontally with a speed of 50 m/s when it hits a block of mass...

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