Question

A 0.149 kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.710 m/s . It has a head-on collision with a 0.308 kg glider that is moving to the left with a speed of 2.27 m/s . Suppose the collision is elastic.1. Find the magnitude of the final velocity of the 0.157kg glider.

2. Find the magnitude of the final velocity of the 0.306kg glider.

Answer 1

v1 = −2.201946 m/s ( to the left)

v2 = 0.7780534 m/s ( to the right)

Step-by-step explanation:

Given the following :

Mass of first glider (m1) = 0.149kg

Initial Speed of first glider (u1) = 0.710 m/s

Mass of second glider (m2) = 0.308kg

Initial Speed of second glider (u2) = 2.27m/s

For elastic collision:

m1u1 + mu2u2 = m1v1 + m2v2

Where V1 and v2 = final velocities if the body after collision.

Taking right as positive ; left as negative

u1 = 0.710m/s ; u2 = - 2.27m/s

u1 - u2 = - (v1 - v2)

0.710 - - 2.27 = - v1 + v2

v2 - v1 = 2.98 - - - - (1)

From:

m1u1 + mu2u2 = m1v1 + m2v2

(0.149 * 0.710) + ( 0.308 * - 2.27) = (0.149 * v1) + (0.308 * v2)

0.10579 + (-0.69916) = 0.149 v1 + 0.308v2

−0.59337 = 0.149 v1 + 0.308v2

Dividing both sides by 0.149

v1 + 2.067v2 = −0.59337 - - - - - (2)

From (1)

v2 = 2.98 + v1

v1 + 2.067(2.98 + v1) = −0.59337

v1 + 6.16 + 2.067v1 = −0.59337

3.067v1 = −0.59337 - 6.16

3.067v1 = −6.75337

v1 = −6.75337 / 3.067

v1 = −2.201946 m/s ( to the left)

From v2 = 2.98 + v1

v2 = 2.98 + (-2.201946)

v2 = 0.7780534 m/s ( to the right)