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1 year ago

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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?

1 answer:

Mice21 [21]1 year ago

8 0

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

$\theta f = 26.3 \deg$
vf/vA= 0.94
2vf^2 / 5vA^2 = 0.35

<h3>What is Collison equilibrium?</h3>

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

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

$\theta f = 26.3 \deg$

(b)

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

vf/vA= 0.94

(c)

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

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

$\theta f = 26.3 \deg$

vf/vA= 0.94

2vf^2 / 5vA^2 = 0.35

Read more about collision theory

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