The momentum of the bumper car after the collision is 1,040 kgm/s.
<h3>
Momentum of bumper</h3>
The change in momentum of the bumper is calculated as follows;
P = v(m1 + m2)
P = 8(55 + 75)
P = 1,040 kgm/s
The momentum of the bumper car after the collision is 1,040 kgm/s.
The direction is still the same.
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Answer:

Explanation:
Puck A's initial speed is
and move in a direction
after the collision.
#
since there's no external force on the system(
#The collision equation can be written as;

The kinetic energies before and after the collision are expressed as:


Let +x along A's initial direction and +y along A's final direction makes the angle 

#Substitute in
:


Hence, the the speed of puck A after the collision is 12.891 m/s
#. The velocity of A after the collision is;

Substitute
into
:

This is the velocity of puck B after the collision, it's speed is:

The velocity of puck B after the collision is 6.8544 m/s
c. The direction of puck B after the collision is:

Hence, the direction of B's velocity after the collision is 62°