A glider of mass 0.160 kg is moving to the right on a frictionless, horizontal air track with a speed of 0.900 m/s . It has a head-on collision with a glider 0.310 kg that is moving to the left with a speed of 2.18 m/s . Suppose the collision is elastic. Find the final velocity (magnitude and direction) of each glider.

- V_a2 is 3.156 m/s in the left direction

- V_b2 is 0.076 m/s in the left direction

Explanation:

Since the 2 gliders are moving on a frictionless surface, then there is no net external force on the system and thus the collision is elastic;

In elastic collisions, the relative velocities before and after collision have the same values.

Thus,

V_a1 - V_b1 = V_b2 - V_a2

Where,

V_a1 is initial velocity of first glider

V_a2 is final velocity of first glider

V_b1 is initial velocity of second glider

V_b2 is final velocity of second glider

From the question,

V_a1 = 0.9 m/s

V_b1 = 2.18 m/s

Thus,

0.9 - (-2.18) = V_b2 - V_a2

V_b2 - V_a2 = 3.08 m/s - - - - (1)

Now, the collision equation is written as;

m_a•V_a1 + m_b•V_b1 = m_a•V_a2 + m_b•V_b2

We are given that;

m_a = 0.16kg

m_b = 0.31 kg

Thus, plugging in the values,

(0.16 x 0.9) + (-0.31 x 2.18) = 0.16•V_a2 + 0.31•V_b2

0.144 - 0.6758 = 0.16•V_a2 + 0.31•V_b2

0.16•V_a2 + 0.31•V_b2 = -0.5318

Divide through by 0.16;

V_a2 + 1.9375•V_b2 = -3.324 - - - (2)

Let's add eq1 and 2 to get;

V_b2 + 1.9375•V_b2 = 3.08 - 3.324

2.9375V_b2 = -0.224

V_b2 = -0.224/2.9375 = - 0.076 m/s

Negative sign means it is in an opposite direction which in this case is to the left.

Thus, V_b2 is 0.076 m/s in the left direction

Now, let's find V_b2.

Let's put -0.076 m/s for V_b2 in eq 1;

Thus,

-0.076 m/s - V_a2 = 3.08 m/s

V_a2 = -0.076 m/s - 3.08 m/s

V_a2 = -3.156 m/s

Negative sign means it is in an opposite direction which in this case is to the left.

Thus, V_a2 is 3.156 m/s in the left direction

d eras and periods

the answer is b, have negligible mass