Collisions and the Speed of Two Pucks

[bowbe]

Homework Statement

A hockey puck moving along the +x axis at 0.5 m/s collides into another puck that is at rest. The pucks have equal mass. The first puck is deflected 37degrees below the +x axis and moves off at 0.36 m/s. Find the speed and direction of the second puck after the collision.

Homework Equations

1/2*m_a*v_ai^2 + 1/2*m_b*v_bi^2 = 1/2*m_a*v_af^2 + 1/2*m_b*v_bf^2
m_a*v_ai + m_b*v_bi = m_a*v_af + m_b*v_bf

The Attempt at a Solution

For part1, I am using the condensed formula of:
v_ai^2 = v_af^2 + v_bf^2
since the masses are equal and the second puck starts at rest, which gives me
v_bf = .3496 (says it is incorrect)

For part2, I know to separate the x-component and y-component to get (condensed again):
x:
0.5 = (.36)(cos37) + (v_bf)(cos[unknown])
y:
0 = (.36)(sin37) + (v_bf)(sin[unknown])

I could solve for the unknown angle if I knew v_bf.

Can anyone help me find what I am doing wrong?

 

[alphysicist]

Hi bowbe,

Homework Statement

A hockey puck moving along the +x axis at 0.5 m/s collides into another puck that is at rest. The pucks have equal mass. The first puck is deflected 37degrees below the +x axis and moves off at 0.36 m/s. Find the speed and direction of the second puck after the collision.

Homework Equations

1/2*m_a*v_ai^2 + 1/2*m_b*v_bi^2 = 1/2*m_a*v_af^2 + 1/2*m_b*v_bf^2

Did you give the whole problem above? This equation is for elastic collisions (kinetic energy conserved). Is this an elastic collision?

m_a*v_ai + m_b*v_bi = m_a*v_af + m_b*v_bf

The Attempt at a Solution

For part1, I am using the condensed formula of:
v_ai^2 = v_af^2 + v_bf^2
since the masses are equal and the second puck starts at rest, which gives me
v_bf = .3496 (says it is incorrect)

For part2, I know to separate the x-component and y-component to get (condensed again):
x:
0.5 = (.36)(cos37) + (v_bf)(cos[unknown])
y:
0 = (.36)(sin37) + (v_bf)(sin[unknown])

Here you have two equations with two unknowns, and so you can solve for both unknowns from just these two. (However, I would write this with one of these terms being negative.)

 

[nlightner]

I would approach this problem by first identifying the relevant equations and principles that govern the motion of objects in collisions. In this case, we can use the conservation of momentum and the conservation of kinetic energy to solve for the unknown variables.

Using the conservation of momentum equation, we can write:

m_a*v_ai + m_b*v_bi = m_a*v_af + m_b*v_bf

Since the masses are equal and the second puck starts at rest, we can simplify this equation to:

v_ai = v_af + v_bf

Next, we can use the conservation of kinetic energy equation to write:

1/2*m_a*v_ai^2 + 1/2*m_b*v_bi^2 = 1/2*m_a*v_af^2 + 1/2*m_b*v_bf^2

Substituting in the expression for v_ai from the first equation, we get:

1/2*m_a*(v_af + v_bf)^2 + 1/2*m_b*v_bi^2 = 1/2*m_a*v_af^2 + 1/2*m_b*v_bf^2

Simplifying and solving for v_bf, we get:

v_bf = (v_af - v_bi)*m_a / m_b

Plugging in the known values, we get:

v_bf = (0.36 - 0.5) * m_a / m_b = -0.14 * m_a / m_b

Since we know that the masses are equal, we can simplify this to:

v_bf = -0.14 * v_af

Plugging in the known value for v_af, we get:

v_bf = -0.14 * 0.36 = -0.0504 m/s

So, the final speed of the second puck after the collision is 0.0504 m/s in the opposite direction of the first puck (along the -x axis). This makes sense intuitively, since the first puck is initially moving faster and in the opposite direction, so it imparts some of its momentum to the second puck, causing it to move in the opposite direction with a slower speed.

I hope this helps! It's always important to approach problems like this with a clear understanding of the relevant principles and equations, and to check your work to make sure it makes sense intuitively.