mₐ vᵢₐ + mᵦ vᵢᵦ = mₐ vfₐ + mᵦ vfᵦ ==> [i and f
means initial and final velocities]
0.5 mₐ vᵢₐ² + 0.5 mᵦ vᵢᵦ² = 0.5 mₐ vfₐ² + 0.5 mᵦ vfᵦ²
The collision is completely specified given the two initial
velocities and the masses of the two objects. Combining the two
equations above gives us a solution to the final velocities for an elastic
collision of two objects:
vfₐ = [(mₐ - mᵦ) vᵢₐ + 2 mᵦ vᵢᵦ]/[mₐ + mᵦ]
vfᵦ = [2 mₐ vᵢₐ − (mₐ - mᵦ) vᵢᵦ]/[mₐ + mᵦ]
substituting the values
mₐ = 70 kg
mᵦ = 0.280 kg
vᵢₐ = 0 m/s
vᵢᵦ = 45.5 m/s
=> vfₐ = [(70 - 0.280) 0 + 2(0.280) 45.5]/[70 + 0.28]
= 1.35 m/s ------------------->(in the original direction
of the puck)
and
=> vfᵦ = [0 − (70 - 0.280) 45.5]/[ 70 + 0.28] = -45.14 m/s
(opposite direction)
Final speeds are therefore:
goalie = 1.35 m/s
puck = 45.14 m/s (opposite direction)
The final velocities of the goalie and the puck after the elastic collision are 0.4 m/s and 45.1 m/s respectively.
The given parameters:
The final velocities of the goalie and the puck after the elastic collision are calculated by applying the principle of conservation of linear momentum as follows;
[tex]m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\\\\70(0) + 0.28(45.5) = 70v_1+ 0.28(-v_2)\\\\12.74 = 70v_1 - 0.28v_2 \ ---(1)[/tex]
Apply one-direction velocity equation;
[tex]u_1 + v_1 = u_2 + v_2\\\\0 + v_1 = 45.5 - v_2\\\\v_1 = 45.5- v_2 \ \ ---(2)[/tex]
Substitute the value of v1 into equation (1);
[tex]12.74= 70(45.5 - v_2)- 0.28v_2\\\\12.74 = 3185 - 70v_2 - 0.28v_2\\\\-3172.26= -70.28v_2\\\\v_2 = \frac{3172.26}{70.28} \\\\v_2 = 45.1 \ m/s[/tex]
[tex]v_1 = 45.5 - v_2\\\\v_1 = 45.5 - 45.1\\\\v_1 = 0.4 \ m/s[/tex]
Thus, the final velocities of the goalie and the puck after the elastic collision are 0.4 m/s and 45.1 m/s respectively.
Learn more about elastic collision here: https://brainly.com/question/7694106
