Question

The eight ball, which has a mass of m = 0.5 kg, is initially moving with a velocity v = 2.5i m/s. It strikes the six ball inelastically, which has an identical mass and is initially at rest. After the collision the eight ball is deflected by an angle of θ = 28° and the six ball is deflected by an angle of Φ = 10.5°, as shown in the figure.

Answer 1

Concept:

A collision that is fully elastic is one in which there is no kinetic energy lost during the contact. In an inelastic collision, some of the kinetic energy is converted into another kind of energy during the contact.

Given:

The mass of the balls is 0.5 kg and the initial velocity of the eight ball is 2.5i  m/s. The six ball is initially at rest. After the collision, the six ball is deflected at an angle Φ = 10.5° and the eight ball is deflected by an angle of θ = 28°.

Find:

The mass of the balls, m = 0.5kg

The initial velocity of the eight ball, v = 2.5i m/s

The eight ball was deflected by an angle after the collision, θ = 28°

The eight ball was deflected by an angle after the collision, Φ = 10.5°

Therefore, the moment is conserved after the collision:

$m\vec{v}=m_6\vec{v_{6}}+m_8\vec{v_{8}}$

Now, $m=m_6=m_8$

$m\vec{v}=m_6\vec{v_6}+m_8\vec{v_8}\\m\vec{v}=m_6\vec{v_6}+m_8\vec{v_8}\\\vec{v}=\vec{v_6}+\vec{v_8}\\vi+0j=v_6(cos(\phi))i+v_6(sin(\phi))j+v_8(cos(\theta))i-v_8(sin(\theta))j\\2.5i=[v_6(cos(\phi))+v_8(cos(\theta))]i+[v_6(sin(\phi))-v_8(sin(\theta))]j$

By comparing the vector of both sides,

$2.5=v_6[cos(\phi)]+v_8[cos(\theta)]$ and;

$0=v_6[sin(\phi)]-v_8[sin(\theta)]\\v_6[sin(\phi)]=v_8[sin(\theta)]\\v_8=\frac{v_6[sin(\phi)]}{sin(\theta)}$

Substituting this value,

$v=v_6[cos(\phi)]+\frac{v_6[sin(\phi)]}{sin(\theta)}cos(\theta)$

$v_6=\frac{v}{cos(\phi)+\frac{sin(\phi)}{sin(\theta)}cos(\theta) }$

$v_6=\frac{2.5}{cos(10.5)+\frac{sin(10.5)}{sin(28)}cos(28) }$

$v_6=\frac{2.5}{0.98+\frac{0.18}{0.469}(0.88) }$

$v_6=\frac{2.5}{0.98+0.337}\\ v_6=1.898\text{m/s}$

The velocity of the eight ball after the collision is:

$v_8=\frac{v_6(sin(\phi)}{sin(\theta)}$

$v_8=\frac{(1.898)(sin(10.5))}{sin(28)}$

$v_8=\frac{(1.898)(0.18)}{0.469}$$v_8=0.728\text{m/s}$

The velocity of the eight ball and the six balls after the collision is 0.728 m/s and 1.898 m/s respectively.

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