Ball A (1.55 kg) moving right at 8.76 m/s makes a head-on collision with ball B (0.752 kg) moving left at 11.4 m/s. After, ball B moves right at 9.03 m/s. What is the final velocity of ball A?

Answer 1

Hello Tamia!

As a general rule of thumb, whenever you see "head-on collision" you must always think of using conservation of momentum! Here, I have Ma as the mass of car A, and Va_i as the initial velocity of car A, etc.

MaVa_i + MbVb_i = MaVa_f + MbVb_f

If we rearrange the equation above, we can solve for the final velocity of car A, Va_f. I will define moving to the right as positive and to the left as negative.

Vb_f = [MaVa_i + MbVb_i - MbVb_f] / Ma

All that's left to do is plug in the numbers and we have our answer! Watch your negatives!

Vb_f = [ (1.55 x 8.76) + (0.752 x -11.4) - (0.752 x 9.03)] / 1.55 = -1.15 m/s

The final velocity of ball A will be 1.15 m/s to the left.

Answer 2

The total momentum of the system before collision is 1.55 kg x 8.76 m/s - 0.752 kg x 11.4 m/s. This is equal to the total momentum after the collision which is 1.55 kg x V_a + 0.752 x 9.03 m/s. Momentum to the right could be considered positive, to the left negative. Equate the momenta before and after collision.

1.55 kg x 8.76 m/s - 0.752 kg x 11.4 m/s. = 1.55 kg x V_a + 0.752 x 9.03 m/s

Solve for V_a.

V_a = -1.15 m/s (to the left)