Answer:
A) U_max = 3 J ; V_a = 0.5 m/s and V_b = 0.5 m/s
B) V_a = 0.25 m/s
V_b = 2.25 m/s
Explanation:
The collision is an elastic collision and thus the energy and momentum are conserved.
We are given;
Mass of Block A; m_a = 2 kg
Mass of Block B; m_b = 6 kg
Initial velocity of Block A; v_ai = 2 m/s
Initial velocity of Block B; v_bi = 0 m/s
A) Thus, total initial kinetic energy is;
KE_initial = ½m_a•v_ai² + ½m_b•v_bi²
Thus;
KE_initial = (½ × 2 × 2²) + (½ × 6 × 0)
KE_initial = 4 J
Now, since they stick together after collision, we can calculate the velocity with which they move together from;
m_a•v_ai + m_b•v_bi = (m_a + m_b)v_2
Where v_2 is the velocity with which they move after sticking together.
v_2 = (m_a•v_ai + m_b•v_bi)/(m_a + m_b)
Plugging in the relevant values, we have;
v_2 = ((2 × 2) + (6 × 0))/(2 + 6)
v_2 = 4/8
v_2 = 0.5 m/s
Final Kinetic energy is;
KE_2 = ½(m_a + m_b) × (v_2)²
KE_2 = ½(2 + 6) × 0.5²
KE_2 = 1 J
Thus, maximum energy stored in the spring bumpers is;
U_max = KE_initial - K_final
U_max = 4 - 1
U_max = 3 J
Since objects A & B, stick together after collision, they will both move with velocity v_2
Thus, at that point, V_a = 0.5 m/s and V_b = 0.5 m/s
B) For elastic collision that's head on, we know that;
Relative velocity of approach = Relative velocity of
separation
Thus;
2 - 0 = V_b - V_a
V_b - V_a = 2
V_b = V_a + 2
From conservation of momentum;
m_a•V_a + m_b•V_b = (m_a + m_b)v_2
We saw earlier that V_b = V_a + 2
Thus;
m_a•V_a + m_b(V_a + 2) = (m_a + m_b)v_2
Plugging in the relevant values;
2(V_a) + 6(V_a + 2) = (2 + 6) × 0.5
2V_a + 6V_a + 2 = 8 × 0.5
8V_a + 2 = 4
8V_a = 4 - 2
8V_a = 2
V_a = 2/8
V_a = 0.25 m/s
From earlier, we saw that;
V_b = V_a + 2
Thus;
V_b = 0.25 + 2
V_b = 2.25 m/s
