Question

Chapter 8: A 0.150kg glider is moving to the right (+x) on a frictionless, horizontal air track with a speed of 0.80m/s. It has an elastic collision with a 0.300kg glider moving to the left (-x) with a speed of 2.20m/s.
a.) What is the initial momentum of each glider? Express the momentum in terms of unit vectors.
b.) Use the relative velocity formula to find v2f in terms of v1f.
c.) Use the relative velocity result to solve conservation of momentum to find the velocity (magnitude and direction) of each glider after the collision.
d.) Show that kinetic energy is conserved.

Answer 1

because the momentum of a mass is \[p=mv\]
a. \[p_1=mv_1\hat{x}=(0.15\cdot 0.8)\hat{x}=0.12\hat{x}[J]\]
by the same consideration
\[p_2=-0.66\hat{x}[J]\]

b. I didn't quite understand the question... we need the relative initial velocity? if so :
\[v_1-v_2=0.8+2.2=3[\frac{m}{s}]\]
so \[\vec{v_2}=\vec{v_1}+3[\frac{m}{s}]\]
c. we need to use two equations.
the first :\[m_1\vec{v_1}+m_2\vec{v_2}=m_1\vec{u_1}+m_2\vec{u_2}\]


the second \[\frac{1}{2}m_1\vec{v_1}^2+\frac{1}{2}m_2\vec{v_2}^2=\frac{1}{2}m_1\vec{u_1}^2+\frac{1}{2}m_2\vec{u_2}^2\]

and in numbers: \[-0.54=0.15\vec{u_1}+0.3\vec{u_2}\]
\[0.72=0.075\vec{u_1}^2+0.15\vec{u_2}^2\]
we get:
\[\vec{u_1}=-3.07[\frac{m}{s}]\]
\[\vec{u_2}=-0.262[\frac{m}{s}]\]
d.
the second equation we used is \[\frac{1}{2}m_1\vec{v_1}^2+\frac{1}{2}m_2\vec{v_2}^2=\frac{1}{2}m_1\vec{u_1}^2+\frac{1}{2}m_2\vec{u_2}^2\]
and it shows that the initial kinetic energy is the same as the kinetic energy after the collision , therefore energy is conserved