Two identical hockey pucks collide on level, frictionless ice. Puck A is going e

Question

Two identical hockey pucks collide on level, frictionless ice. Puck A is going east at 4m/s. Puck B is going west at 2m/s. They strike one another in a glancing fashion, so that puck B goes north after the collision, moving at a speed of 2.83 m/s.

Find the velocity of puck A after the collision.

What I know from my own knowledge:
I know how to figure out the problems if they give you two out of the three variables but how do you do a problem when they only give you one variable (velocity)? I drew out the problem and labeled everything and I know that the Sum of P_i=the sum of P_f so . . . P_Ai+P_Bi=P_Af+P_Bf, which then equals: (mAivAi)+(mBivBi)=(m_Af)(v_Af)+(m_Bf)(v_Bf). I think I'm supposed to somehow find the mass first and then do the rest of the equation. Since they are two identical hockey pucks, if I get the mass for one hockey puck, I'll have the mass for the second too I think.

Anyway, help please :D. I have a feeling it's easier than I'm trying to make it.

Explanation / Answer

In horizontal direction conserve momentum
so m*4 - m*2 = m*vx
vx = 2 m/s
in vertical direction
m*2.83 + m*vy = 0
vy = -2.83 m/s
Velocity is 2i - 2.83j
(i is towards east and j is towards north)
speed = (vx^2+vy^2) = 3.465 m/s

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.