a) what is V 2fy ? V 2fx ? then b)what is V 1fy ? V 1fx? c) final speed of objec

Question

a) what is V2fy? V2fx?

then

b)what is V1fy? V1fx?

c) final speed of object 1? d)What is the angle from the diagram? (in degrees)

The diagram shows a two-dimensional collision for which we will assume momentum is conserved. The objects have equal masses. Initially, object 1 is moving in the positive x direction at a speed of 10.5 m/s, while object 2 is at rest. After the collision, object 2 is moving at a speed of 6.4 m/s and in a direction 37.6 above the positive x axis. The final velocity of object 1 makes an angle with the positive x direction, as shown After 2f Before 37.6 +x V1i 2i V1f Determine the magnitude and direction of the final velocity of object 1. We will use Equations 7.5.2, the conservation of momentum equations in component form

Explanation / Answer

In x direction, Total Initial Momentum, px = m*v1i
In y direction, Total Initial Momentum, py = 0

Final Momentum in X dircetion,
pxf = m*V2f*cos(37.6) + m*V1f*cos()


Final Momentum in Y dircetion,
pyf = m*V2f*sin(37.6) - m*V1f*sin()

Gven,
v1f = 10.5 m/s
v2f = 6.4 m/s

Using Momentum Conservation,
Initial Momentum = Final momentum

In X axis,
Px = Pxf
m*v1i = m*V2f*cos(37.6) + m*V1f*cos()
10.5 = 6.4*cos(37.6) + V1f*cos()
V1f*cos() = 5.43   -------1

In y axis,
Py = Pyf
m*V2f*sin(37.6) - m*V1f*sin() = 0
V1f*sin() =  6.4*sin(37.6)
V1f*sin() = 3.9  --------2

Dividing 2/1
tan() = 3.9/5.43
= 35.7

(a)
V2fy = 6.4*sin(37.6) = 3.9 m/s
V2fx = 6.4*cos(37.6) = 5.07 m/s

(b)
V1fy = 5.43 m/s
V1fx =  3.9 m/s

(c)
Final Speed of object1,
V1 = 6.68 m/s
(d)
Angle from the diagram, = 35.7o

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