Elastic collision with Conservation of momentum problem?

[nchin]

elastic collision with Conservation of momentum problem?

i need help with part b.

Two titanium spheres approach each other head-on with the same speed and collide elastically, After the collision, one of the spheres, whose mass (m1) is .3 kg, remains at rest.

(a) What is the mass of the other sphere?
(b) Assume that the initial speed of each sphere was 2.0 m/s. What is the speed of the two-sphere center of mass?

a)
I know i did part a correctly. here's what i did:

momentum is conserved: m1 * u - m2 * u = m2 * v
or (m1 - m2) * u = m2 * v
Also, for an elastic head-on collision, we know that the
relative velocity of approach = relative velocity of separation
(from conservation of energy), or, for this problem,
2u = v
Then
(m1 - m2) * u = m2 * 2u
m1 - m2 = 2 * m2
m1 = 3 * m2
m1 is the sphere that remained at rest (hence its absence from the RHS), so
m2 = 0.3kg / 3
m2 = 0.1 kg

b) this part confuses me, here's what i did

(m1 - m2) * u = m2 * v
(.3kg - .1kg)(2.0m/s) = .1kg * v
.4 kg = .1 v
v = 4 m/s

What my teacher did:
(.3g - .1g) * 2.0m/s = (.3g + .1g) * v

I understand the left hand side but i don't get the right hand side. Why is m1 added to m2 when m1 is at rest which makes its v = zero??

v = +1.00m/s

since the answer is positive, what does that mean? Also, if v was -1.00m/s what would that mean?

thanks!

 

[Saitama]

Your teacher used this:
$$v_{CM}=\frac{m_1v_1+m_2v_2}{m_1+m_2}$$
$v_{CM}$ is the velocity of center of mass.

 

[nchin]

Your teacher used this:
$$v_{CM}=\frac{m_1v_1+m_2v_2}{m_1+m_2}$$
$v_{CM}$ is the velocity of center of mass.

I know he uses that formula but shouldn't m1 be zero since its final v is zero because it's at rest?

 

[Saitama]

I know he uses that formula but shouldn't m1 be zero since its final v is zero because it's at rest?

If you carefully look at the formula i posted, the numerator is the linear momentum of the system. Here, since the momentum is conserved, the numerator does not change. Your teacher used this and found momentum of system before collision which saved him from doing more steps. Even if you find the final velocities after the collision and substitute it in the formula, you get the same answer as your teacher.

 

[Nickie]

First of all, great job on part a! Your understanding of conservation of momentum is correct.

For part b, your teacher's approach is also correct. Let's break down the right hand side of the equation:

(.3kg + .1kg) * v

= (m1 + m2) * v

= (m1 + m2) * 2u

= (m1 + m2) * 2 * (u/2)

= (m1 + m2) * vcm

Where vcm is the velocity of the center of mass. This equation uses the fact that in an elastic collision, the relative velocity of approach is equal to the relative velocity of separation. So, we can say that the initial speed of each sphere (2u) is equal to the final speed of the center of mass (vcm).

Now, let's look at the answer. Since the final speed of the center of mass is 1.00 m/s, this means that the two spheres are moving together at a speed of 1.00 m/s after the collision. This makes sense because the larger sphere (m1) was initially at rest, so it must have gained some velocity from the smaller sphere (m2) in order to move together after the collision.

If the final speed of the center of mass was -1.00 m/s, this would mean that the two spheres are moving in opposite directions after the collision. This would happen if the smaller sphere (m2) had a larger initial velocity than the larger sphere (m1), causing them to separate after the collision.

Overall, the positive or negative sign of the final speed of the center of mass tells us the direction of the motion of the two spheres after the collision. A positive value means they are moving together, while a negative value means they are moving in opposite directions.