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

momentum of soft ball three times less than baskey ball

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft