Answer:
(a) The x-component of the second sphere's velocity is 0.775 m/s
(b) The y-component of the second sphere's velocity is -0.338 m/s
Explanation:
Conservation of Linear Momentum
The principle of the conservation of linear momentum states that the total momentum of a closed system of particles is conserved while no external forces act on any part of it, regardless of the interaction between the particles. The momentum is a vector, thus we must analyze both axes in the calculations.
The momentum can be computed as
[tex]\vec p=m.\vec v[/tex]
where m is the mass of the particle, and [tex]\vec v[/tex] is its velocity. In a system of two particles, the total initial momentum is
[tex]\vec p_o=m_1.\vec v_1+m_2.\vec v_2[/tex]
And the final momentum is
[tex]\vec p_1=m_1.\vec v_1'+m_2.\vec v_2'[/tex]
Since the total momentum is conserved
[tex]m_1.\vec v_1'+m_2.\vec v_2'=m_1.\vec v_1+m_2.\vec v_2[/tex]
According to the conditions of the problem, both masses are identical and sphere 2 is initially at rest (v2=0), thus
[tex]m.\vec v_1'+m.\vec v_2'=m.\vec v_1[/tex]
Simplifying by m
[tex]\vec v_1'+\vec v_2'=\vec v_1\text{ ......[1]}[/tex]
We know that the first sphere has a velocity of 1.5 m/s to the right. This means that the vertical component of v1 is 0:
[tex]\vec v_1=<1.5,0>[/tex]
We also know after the collision this same sphere travels at 0.8 m/s at an angle of 25°. The components of this velocity are
[tex]\vec v_1'=<0.8cos25^o,0.8sin25^o>[/tex]
(a) Solving the equation [1] for v2
[tex]\vec v_2'=\vec v_1-\vec v_1'[/tex]
[tex]\vec v_2'=<1.5,0>-<0.8cos25^o,0.8sin25^o>[/tex]
[tex]\vec v_2'=<1.5,0>-<0.725,0.338>[/tex]
[tex]\vec v_2'=<0.775,-0.338>[/tex]
The x-component of the second sphere's velocity is 0.775 m/s
(b) The y-component of the second sphere's velocity is -0.338 m/s
