A 0.500 kg bead slides on a straight frictionless wire with a velocity of 1.50 cm/s to the right. The bead collides elastically with a larger 0.750 kg bead initially at rest. After the collision, the smaller bead moves to the left with a velocity of 0.70 cm/s. Find the distance the larger bead moves along the wire in the first 2.0 s following the collision.

We've never done a problem like this in class. I tried finding the velocity of the other bead and plugging it into a kinematic equation but the answer was wrong.

why does V*2 = distance?

Because he found the velocity per second so you have to multiply by two to get the answer for two seconds

well, you do indeed find the velocity of the large bead, but, you have to then find the distance it travels in 2 seconds.

Momentum
initial=final
.5*1.5=.5(-.70)+7.50(V)

V= (.75+.35)/7.50 cm/s

distance= V*2

check my math, I did it in my head.

To solve this problem, we can first consider the conservation of momentum and the conservation of kinetic energy in an elastic collision.

1. Conservation of momentum:
In an isolated system, the total momentum before the collision is equal to the total momentum after the collision. Mathematically, we have:

(m1 * v1) + (m2 * v2) = (m1 * v1') + (m2 * v2')

where m1 is the mass of the smaller bead, v1 is its initial velocity, m2 is the mass of the larger bead, v2 is its initial velocity, and v1' and v2' are their velocities after the collision.

Substituting the given values:
(0.500 kg * 1.50 cm/s) + (0.750 kg * 0 cm/s) = (0.500 kg * -0.70 cm/s) + (0.750 kg * v2')

Simplifying the equation:
0.750 kg * v2' = 0.750 kg * 1.50 cm/s + 0.500 kg * 1.50 cm/s
v2' = (0.750 kg * 1.50 cm/s + 0.500 kg * 1.50 cm/s) / 0.750 kg

By calculating this, we find v2' = 3 cm/s.

2. Conservation of kinetic energy:
In an elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Mathematically, we have:

(0.5 * m1 * v1^2) + (0.5 * m2 * v2^2) = (0.5 * m1 * v1'^2) + (0.5 * m2 * v2'^2)

Substituting the given values:
(0.5 * 0.500 kg * (1.50 cm/s)^2) + (0.5 * 0.750 kg * (0 cm/s)^2) = (0.5 * 0.500 kg * (-0.70 cm/s)^2) + (0.5 * 0.750 kg * (3 cm/s)^2)

Simplifying the equation:
(0.5 * m1 * v1'^2) + (0.5 * m2 * v2'^2) = (0.5 * 0.500 kg * (-0.70 cm/s)^2) + (0.5 * 0.750 kg * (3 cm/s)^2)

By calculating this, we find (0.5 * m1 * v1'^2) + (0.5 * m2 * v2'^2) = 0.752 J.

Now, to find the distance the larger bead moves along the wire in the first 2.0 s following the collision, we need to determine its velocity.

By using the equation of motion:
v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time,

we can rearrange the equation to solve for the initial velocity:

u = v - at

Given that the initial velocity is 0 cm/s, and the time is 2.0 s, we can calculate the acceleration:

a = (v - u) / t

Substituting the values:
a = (3 cm/s - 0 cm/s) / 2.0 s

By calculating this, we find a = 1.5 cm/s^2.

Now, we can calculate the distance using the equation of motion:

s = ut + (1/2)at^2

Substituting the values:
s = (0 cm/s)(2.0 s) + (1/2)(1.5 cm/s^2)(2.0 s)^2

By calculating this, we find s = 3 cm.

Therefore, the distance the larger bead moves along the wire in the first 2.0 s following the collision is 3 cm.