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tester [92]
3 years ago
7

a body of mass 1.5kg, traveling along the positive x axis with speed 4.5m/s,collides with another body B of mass 3.2kg which,ini

tially is at rest. A is deflected and moves with a speed of 2.1m/s in a direction which is 30 degrees below the x axis. B Is set in motion at angle b above the x axis. calculate the velocity of B after collision.,
Physics
1 answer:
Yuki888 [10]3 years ago
4 0
Data:
m₁ = 1.5kg
m₂ = 3.2kg
α = -30° (negative because it is below the x-xis)
v_{1i} = initial speed of object 1 = 4.5m/s
v_{2i} = <span>initial speed of object 2 = 0m/s
v_{1f} = final</span><span> speed of object 1 = 2.1m/s
v_{2f} = ?
</span>β = ?

Since the motion after the collision is in 2 dimentions, it is better to write the speeds with their components along the x and the y-axis:
v_{1ix} = initial speed of object 1 along x-axis = 4.5m/s
v_{1iy} = <span>initial speed of object 1 along y-axis = 0m/s
</span>v_{2ix}<span> = initial speed of object 2 along x-axis = 0m/s
v_{2iy} = </span><span>initial speed of object 2 along y-axis = 0m/s
</span>
v_{1fx} = final speed of object 1 along x-axis = 2.1 cos(-30) = 1.82m/s
v_{1iy} = final speed of object 1 along y-axis = 2.1 sin(-30) = -1.05m/s 

In this kind of collision, we have the conservation of momentum, therefore we can write the system:
\left \{ {{m_{1} v_{1ix} + m_{2} v_{2ix} =  m_{1} v_{1fx} + m_{2} v_{2fx} } \atop { m_{1} v_{1iy} + m_{2} v_{2iy} =  m_{1} v_{1fy} + m_{2} v_{2fy}}} \right.

Considering the terms that are zero, it becomes:
\left \{ {{m_{1} v_{1ix} = m_{1} v_{1fx} + m_{2} v_{2fx} } \atop {0 = m_{1} v_{1fy} + m_{2} v_{2fy}}} \right.

Let's face first the y-component:
m_{2} v_{2fy} = -m_{1} v_{1fy}

therefore:
v_{2fy} = \frac{-m_{1} v_{1fy}}{m_{2}}=\frac{-(1.5)(-1.05)}{3.2} = 5.04m/s

Now, let's face the x-component:
v_{2fx}=\frac{m_{1} v_{1ix} - m_{1} v_{1fx}}{m_{2}} =
\frac{m_{1} (v_{1ix} - v_{1fx})}{m_{2}} = \frac{(1.5)(4.5-1.82)}{3.2} = 1.26m/s

Now that we have the two components, we can find:

v_{2f} = \sqrt{ v_{2fx}^2 + v_{2fy}^2 } = \sqrt{5.04^{2}  +  1.26^{2} } = 6.35m/s

Lastly, the angle can be found with trigonometry:

β = tan⁻¹(\frac{ v_{2fy} }{ v_{2fx} }) = tan⁻¹(<span>\frac{ 1.26} }{ 5.04} }) = 14°</span>



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