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3 years ago

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A 22g bullet traveling 210 m/s penetrates a 2.0kg block of wood and emerges going 150m/s. If the block were stationary on a fric

tionless plane before the collision, what is the velocity of the block after the bullet passes through? What is the kinetic energy loss of the system?

1 answer:

Anna007 [38]3 years ago

6 0

Answer:

speed of block is 0.665 m/s

Kinetic energy loss is 237.16 J

Explanation:

Here we can use momentum conservation as there is no external force on the system in horizontal direction

so here we will have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

$0.022(210) + 2(0) = 0.022(150) + 2 v$

$4.62 = 3.3 = 2 v$

$v = \frac{4.63 - 3.3}{2}$

$v = 0.665 m/s$

Now kinetic energy loss in the system is given as

$Loss = KE_i - KE_f$

$Loss = \frac{1}{2}(0.022)(210^2) - (\frac{1}{2}(0.022)(150^2) + \frac{1}{2}(2)(0.665)^2)$

$Loss = 237.16 J$

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2 years ago

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2 years ago

A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turn

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a) The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

Explanation:

Statement is incomplete. The complete description is now described below:

<em>A satellite in outer space is moving at a constant velocity of 21.4 m/s in the y direction when one of its onboard thruster turns on, causing an acceleration of 0.250 m/s2 in the x direction. The acceleration lasts for 45.0 s, at which point the thruster turns off. </em>

<em>(a) What is the magnitude of the satellite's velocity when the thruster turns off</em>

<em>(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis. ° counterclockwise from the +x-axis</em>

Let be x and y-directions orthogonal to each other and the satellite is accelerated uniformly from rest in the +x direction and moves at constant velocity in the +y direction. The velocity vector of the satellite ( $\vec{v}_{S}$ ), measured in meters per second, is:

$\vec{v}_{S} = (v_{o,x}+a_{x}\cdot t)\,\hat{i}+v_{y}\,\hat{j}$

Where:

$v_{o,x}$ - Initial velocity in +x direction, measured in meters per second.

$a_{x}$ - Acceleration in +x direction, measured in meter per square second.

$t$ - Time, measured in seconds.

$v_{y}$ - Velocity in +y direction, measured in meters per second.

If we know that $v_{o,x} = 0\,\frac{m}{s}$ , $a_{x} = 0.250\,\frac{m}{s^{2}}$ , $t = 45\,s$ and $v_{y} = 21.4\,\frac{m}{s}$ , the final velocity of the satellite is:

$\vec{v}_{S} = \left[0\,\frac{m}{s}+\left(0.250\,\frac{m}{s^{2}} \right)\cdot (45\,s) \right]\,\hat{i}+\left(21.4\,\frac{m}{s} \right)\,\hat{j}$

$\vec{v_{S}} = 11.25\,\hat{i}+21.4\,\hat{j}\,\,\left[\frac{m}{s} \right]$

a) The magnitud of the satellite's velocity can be found by the resource of the Pythagorean Theorem:

$\|\vec {v}_{S}\| = \sqrt{\left(11.25\,\frac{m}{s} \right)^{2}+\left(21.4\,\frac{m}{s} \right)^{2}}$

$\|\vec{v}_{S}\| \approx 24.177\,\frac{m}{s}$

The magnitude of the satellite's velocity when the thruster turns off is approximately 24.177 meters per second.

b) The direction of the satellite's velocity when the thruster turns off is determined with the help of trigonometric functions:

$\tan \alpha = \frac{v_{y}}{v_{x}} = \frac{21.4\,\frac{m}{s} }{11.25\,\frac{m}{s} }$

$\tan \alpha = 1.902$

$\alpha = \tan^{-1}1.902$

$\alpha \approx 62.266^{\circ}$

The direction of the satellite's velocity when the thruster turns off is approximately 62.266º.

4 0

3 years ago