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Sergio039 [100]
3 years ago
6

I WIL A 2.00-kg object traveling east at 20.0 m/s collides with a 3.00-kg object traveling west at 10.0 m/s. After the collision

, the 2.00-kg object has a velocity 5.00 m/s to the west. How much kinetic energy was lost during the collision? . .
Physics
1 answer:
pishuonlain [190]3 years ago
4 0
Using the following given values:

Object 1:
Mass = M1 = 2 kg
Velocity before collision = Vb1 = 20 m/s
Velocity after collision = Va1 = -5 m/s 

Object 2:
Mass = M2 = 3 kg
Velocity before collision = Vb2 = -10 m/s
Velocity after collision = Va2 = ? m/s<span> 
</span>
Obtaining Va2 via law of conservation of momentum:

total momentum after collision = total momentum before collision
M1 * Va1 + M2 * Va2 = M1 * Vb1 + M2 * Vb2
2*-5 + 3Va2 = 2*20 + 3*-10
Va2 = 6.67

Total kinetic energy before collision:

KE1 = (1/2)*M1*Vb1^2 + (1/2)*M2*Vb2^2
<span>KE1 = (1/2)*2*(20)^2 + (1/2)*3*(-10)^2
KE1 = 550 J

</span>Total kinetic energy after collision:

KE2 = (1/2)*M1*Va1^2 + (1/2)*M2*Va2^2
<span>KE2 = (1/2)*2*(-5)^2 + (1/2)*3*(6.67)^2
KE2 = 91.73 J
</span>
Total kinetic energy lost:

Energy lost = KE1 - KE2 = 550 - 91.73 = 458.27 J
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Answer:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

Explanation:

For this problem to be easier to calculate, we can represent the triangle as a right triangle whose right angle is located at the origin of a coordinate system. (See picture attached).

With this disposition of the triangle, we can start finding our integral. The hydrostatic force can be set as an integral with the following shape:

\int\limits^a_bγhxdy

we know that γ=62.5 lb/ft^{3}

from the drawing, we can determine the height (or depth under the water) of each differential area is given by:

h=8-y

x can be found by getting the equation of the line, which we'll get by finding the slope of the line and using one of the points to complete the equation:

m=\frac{y_{2}-y_{1}}{x_{2}-x{1}}

when substituting the x and y-values given on the graph, we get that the slope is:

m=\frac{0-6}{7-0}=-\frac{6}{7}

once we got this slope, we can substitute it in the point-slope form of the equation:

y_{2}-y_{1}=m(x_{2}-x_{1})

which yields:

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which simplifies to:

y-6=-\frac{6}{7}x

we can now solve this equation for x, so we get that:

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with this last equation, we can substitute everything into our integral, so it will now look like this:

\int\limits^6_0{(62.5)(8-y)(-\frac{7}{6}y+7)}\,dy

Now that it's all written in terms of y we can now simplify it, so we get:

62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)}dy

we can now proceed and evaluate it.

When using the power rule on each of the terms, we get the integral to be:

62.5[\frac{7}{18}y^{3}-\frac{49}{6}y^{2}+56y]^{6}_{0}

By using the fundamental theorem of calculus we get:

62.5[(\frac{7}{18}(6)^{3}-\frac{49}{6}(6)^{2}+56(6))-(\frac{7}{18}(0)^{3}-\frac{49}{6}(0)^{2}+56(0))]

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62.5\int\limits^6_0 {(\frac{7}{6}y^{2}-\frac{49}{3}y+56)  } \, dy = 7875 lb

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