What Is the Speed and Direction of the Second Hockey Puck After Collision?

[~*Noor*~]

Hello,

Q.) A hockey puck moving at 0.48 m/s collides elastically with another puck that was at rest. The pucks have equal mass. The first puck is deflected 39° to the right and moves off at 0.35 m/s. Find the speed and direction of the second puck after the collision. Make sure the speed is in m/s and direction in terms of degrees

A.) The way i tried solving this problem is below:

m1=m2

Hockey Puck #1:
p=p(final)-p(initial)
=m1(vfinal-vinitial)


Hockey Puck #2:
=m2(vfinal-vinitial)


p1=p2

m1(v1final-v1initial)=m2(v2final-v2initial)

Both the masses cancel out... we have the following:
v1initial=.48
v1final= .35cos39
v2initial=0

Plug in formula and get v2final=-.21 since it's going to the left...but this answer is wrong.

I tried 39° for the second puck since they both weight the same but this is wrong as well.


Please i don't know why I'm getting this wrong. Any help will be appeciated

 

[Kickbladesama]

https://www.physicsforums.com/showthread.php?t=262547

 

[baba89]

A.) Hello, I can help you with this problem. First, let's review some key concepts related to momentum and elastic collisions. Momentum is a property of a moving object, and it is defined as the product of an object's mass and its velocity. In an elastic collision, the total momentum of the system is conserved, meaning that the total momentum before the collision is equal to the total momentum after the collision. In this case, we have two pucks colliding, so the total momentum before the collision is equal to the total momentum after the collision. This can be expressed mathematically as:

m1v1initial + m2v2initial = m1v1final + m2v2final

Since the masses of the two pucks are equal, we can simplify this equation to:

v1initial + v2initial = v1final + v2final

Now, let's plug in the values given in the problem:

v1initial = 0.48 m/s
v2initial = 0 m/s (since the second puck is at rest)
v1final = 0.35 m/s (since the first puck is deflected 39° to the right)
v2final = ? (this is what we are trying to find)

So, the equation becomes:

0.48 + 0 = 0.35cos39 + v2final

Solving for v2final, we get:

v2final = 0.21 m/s to the left

Therefore, the speed of the second puck after the collision is 0.21 m/s and its direction is to the left. It is important to note that the angle of deflection for the second puck is not given in the problem, so we cannot determine it. Also, the direction of the first puck is not relevant in this problem since it is not involved in the collision. I hope this helps clarify the problem for you. Keep up the good work in your studies!