Two balls of equal mass collide and stick together as shown in the figure. The initial velocity of ball B is twice that of ball A. a. Calculate the angle above the horizontal of the motion of mass A + B after the collision. b. What is the ratio of the final velocity of the mass A + B to the initial velocity of ball A, /? c. What is the ratio of the final energy of the system to the initial energy of the system, /? Is the collision elastic or inelastic?

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OKPALAWALTER8GO OKPALAWALTER8GO · Answer 1

The angle above the horizontal of the motion of mass A + B after the collision. as well as others are mathematically given as

According to the collision hypothesis of chemistry, chemical reactions happen when two molecules or atoms collide.

Generally, the equation for Collison equilibrium is mathematically given as

For horizontal motion

mvAx+m2vAx = 2mvfx

For vertical motion

-mvAy+m2vAy = 2mvfy

vfy = 1/2 vAy

[tex]\tan \theta f = vfy/vfx \\\\\tan \theta f = 1/3 vAy/vAx \\\\\tan \theta f = vAy/vAx \\\\\theta f = tan-1 ( 1/3 tan (56) )\\\\[/tex]

[tex]\theta f = 26.3 \deg[/tex]

(b)

[tex]v fy = vf sin \theta f\\\\vAy = vA sin theta i\\\\vf/vA = vfy/vAy sin( \theta i)/sin(\theta f)\\\\vf/vA = 1/2 sin(56)/sin(26.3) \\\\[/tex]

vf/vA= 0.94

(c)

[tex]Ef/Ei = \frac{(1/2 * (2m) * (vf)^2) }{ (1/2 * m * (vA)^2) + 1/2*m*(2vA)^2) ]}[/tex]

2vf^2 / 5vA^2 = 0.35

In conclusion, The angle above the horizontal of the motion of mass A + B after the collision. as well as others

[tex]\theta f = 26.3 \deg[/tex]

vf/vA= 0.94

2vf^2 / 5vA^2 = 0.35