Question

Two identical pucks collide elastically on an air hockey table. Puck 1 was originally at rest; puck 2 has an incoming speed of 6.00 m/s and scatters at an angle of 30° with respect to its incoming direction. What is the velocity (magnitude and direction) of puck 1 after the collision?

a) 3.00 m/s ˆi
b) 5.20 m/s ˆi
c) 3.00 m/s ˆj
d) 5.20 m/s ˆj

Answer 1

When two identical pucks collide elastically, if one puck is at rest, its final velocity after the collision will be 0 m/s.

Step-by-step explanation:

When the two identical pucks collide elastically, the momentum is conserved. Since one puck was originally at rest, its final velocity after the collision will depend on the velocity of the other puck.

To find the velocity of the first puck after the collision, we can use the law of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision. Since one puck is originally at rest, its initial momentum is zero. The formula to calculate the final velocity is:

m1 × v1_final = m2 × v2_final

Where m1 and m2 are the masses of the pucks, and v1_final and v2_final are their final velocities.

We know that the two pucks have identical masses. Let's assume the mass is represented by 'm' for simplification. Since m1 = m2 = m, the equation can be rewritten as:

m × 0 = m × v2_final

Since the mass cancels out, we find that the final velocity of the first puck is 0 m/s.