Ask Question
8 July, 21:08

Two hockey pucks, labeled A and B, are initially at rest on asmooth ice surface and are separated by a distance of18.0 m. Simultaneously, each puck is given a quickpush and they begin to slide directly toward each other. Puck Amoves with a speed of v_A = 3.50 m/s, and puck B moves with a speed ofv_B = 3.90 m/s.

What is the distance that puck A covers prior to the collision?

+4
Answers (1)
  1. 8 July, 21:36
    0
    8.505 m

    Explanation:

    Let V1 and V2 be velocities of puck A and B respectively

    Since A and B move in the same direction, so the relative velocity will be V1+V2=3.5+3.9=7.4m/s

    Or

    Vr=7.4 m/s

    Distance=S = 18 m

    Time = t=?

    S=Vr*t

    ==> t=S/Vr

    ==> t = 18/7.4=2.43 sec

    At this time both will strike together

    Distance by puck A

    V1=3.5 m/s

    Time=t = 2.43 sec

    Distance covered=d=?

    d=V1*t=3.5*2.43=8.505 m

    So, puck A will cover 8.505 meters before collision
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Two hockey pucks, labeled A and B, are initially at rest on asmooth ice surface and are separated by a distance of18.0 m. Simultaneously, ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers