A.5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.
1. determine the direction and speed of the pucks' motion if they stick to each other after the collision.
2. suppose instead that the pucks do not stick together. if the 1.5kg puck moves to the right at .231 m/s after the collision, was the collision elastic? provide evidence!
3. determine the size and direction of the average force acting on the .5kg puck during the second collision, if the collision lasted for 25 ms.

Explanation:

Total momentum of the system before the collision

.5 x 3 - 1.5 x 1.5 = -0.75 kg m/s towards the left

If v be the velocity of the stuck pucks

momentum after the collision = 2 v

Applying conservation of momentum

2 v = -  .75

v =  - .375 m /s

Let after the collision v be the velocity of .5 kg puck

total momentum after the collision

.5 v + 1.5 x .231 = .5v +.3465

Applying conservation of momentum law

.5 v +.3465 = - .75

v = - 2.193 m/s

2 ) To verify whether the collision is elastic or not , we verify whether the kinetic energy is conserved or not.

Kinetic energy before the collision

= 2.25 + 1.6875

=3.9375 J

kinetic energy after the collision

= .04 + 1.2 =1.24 J

So kinetic energy is not conserved . Hence collision is not elastic.

3 ) Change in the momentum of .5 kg

1.5 - (-1.0965 )

= 2.5965

Average force applied = change in momentum / time

= 2.5965 / 25 x 10⁻³

= 103.86 N

do you have a formula possibly