Question

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.

Answer 1

- V_a2 is 3.156 m/s in the left direction

- V_b2 is 0.076 m/s in the left direction

Step-by-step 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

Answer 2

m1 = 0.160kg

v1 = 0.900 m/s before the collision

m2 = 0.310kg

v2 = - 2.18m/s

If final vel. of m1 is v1 and of mass m2 is v2, then

momentum is conserved

m1u1 + m2u2 = m1v1 +m2v2_____________(1)

as kinetic energy is conserved,

(1/2)m1u1^2+(1/2)m2u2^2 = (1/2)m1v1^2+(1/2)m2v2^2__(2)

solving these equations,

v1=[2 m2u2 +(m1-m2) u1] / (m1+m2)

v2 = [2 m1u1 +(m2-m1) u2] / (m1+m2)

substituting values of u1,u2 etc we get

v1 = { 2*0.310*( -2.18) + [( 0.160 -0.310)0.900] } / [ 0.142+0.310]

v1 = - 0.4847m/s

v2 = [2 m1u1 +(m2-m1) u2] / (m1+m2)

= ( 2 × 0.160×0.900+(0.320-0.160)0.310 / (0.160+ 0.310)

v2 = 0.718 m/s

Step-by-step explanation:

The collision is elastic ,meaning both momentum and the kinetic energy remain conserved.we take the velocity towards the right positive and to the left negative.