Blocks A (mass 2.00 kg) and B (mass 6.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 2.00 m>s. The blocks are equipped with ideal spring bumpers, as in Example 8.10 (Section 8.4). The collision is head-on, so all motion before and after the collision is along a straight line. (a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time. (b) Find the velocity of each block after they have moved apart

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

0.863333

86%

hope this

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.