A cue ball (mass = 0.145 kg) is at rest on a frictionless pool table. The ball is hit dead center by a pool stick, which applies an impulse of +1.45 N · s to the ball. The ball then slides along the table and makes an elastic head-on collision with a second ball of equal mass that is initially at rest. Find the velocity of the second ball just after it is struck

Too much work. ALL of the momentum must go to the second ball for perfectly elastic involving same masses(should be a give away that no info is provided post-collision).

FT = mv
1.45/.145 = v
You can verify quickly by making the velocity of the second 10 and the answer pops right out.
Mind you this only works if the masses are the same.

To find the velocity of the second ball just after it is struck by the first ball, we can use the principle of conservation of linear momentum.

According to the principle of conservation of linear momentum, the total momentum before the collision is equal to the total momentum after the collision.

So, we can write:

(mass of cue ball) x (velocity of cue ball) + (mass of second ball) x (velocity of second ball before collision) = (mass of cue ball) x (final velocity of cue ball) + (mass of second ball) x (final velocity of second ball)

Since the cue ball is initially at rest, its initial velocity (before the collision) is zero:

(0.145 kg) x (0) + (0.145 kg) x (velocity of second ball before collision) = (0.145 kg) x (final velocity of cue ball) + (0.145 kg) x (final velocity of second ball)

Since the collision is elastic, the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Since the cue ball is the only object moving with kinetic energy before the collision, the total kinetic energy before the collision is equal to (1/2) x (mass of cue ball) x (velocity of cue ball)^2:

(1/2) x (0.145 kg) x (0)^2 = (1/2) x (0.145 kg) x (final velocity of cue ball)^2 + (1/2) x (0.145 kg) x (final velocity of second ball)^2

0 = (1/2) x (0.145 kg) x (final velocity of cue ball)^2 + (1/2) x (0.145 kg) x (final velocity of second ball)^2

Simplifying this equation, we get:

0 = (0.0725 kg) x (final velocity of cue ball)^2 + (0.0725 kg) x (final velocity of second ball)^2

Since the mass of the cue ball and the second ball are equal, we can solve for the final velocities of each ball independently.

0 = (final velocity of cue ball)^2 + (final velocity of second ball)^2

Now we can substitute the given impulse for the cue ball into the equation.

Impulse = change in momentum

(1.45 N · s) = (mass of cue ball) x (final velocity of cue ball) - (mass of cue ball) x (initial velocity of cue ball)

(1.45 N · s) = (0.145 kg) x (final velocity of cue ball) - (0.145 kg) x (0)

(1.45 N · s) = (0.145 kg) x (final velocity of cue ball)

(final velocity of cue ball) = (1.45 N · s) / (0.145 kg)

(final velocity of cue ball) = 10 m/s

Substituting this value into the previous equation, we get:

0 = (10 m/s)^2 + (final velocity of second ball)^2

(final velocity of second ball)^2 = -100 m^2/s^2

Since velocity cannot be negative, there must be an error in the calculation or in the given values. Please double-check the values given for the impulse and the masses of the cue ball and second ball.

To find the velocity of the second ball just after it is struck, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

1. Calculate the momentum of the cue ball before the collision:
Momentum = mass x velocity
Since the cue ball is at rest, its velocity is zero, so the momentum is also zero.

2. Calculate the momentum of the cue ball after the collision:
The impulse applied to the cue ball by the pool stick is equal to the change in momentum of the cue ball.
Impulse = change in momentum
Impulse = mass x change in velocity
Given impulse = +1.45 N · s and mass = 0.145 kg,
Change in velocity = impulse / mass
Change in velocity = 1.45 N · s / 0.145 kg
Change in velocity = 10 m/s

3. Since the collision is elastic, the momentum of the cue ball is transferred to the second ball.
Therefore, the momentum of the second ball after the collision is also 10 m/s.

4. Calculate the velocity of the second ball after the collision:
Velocity = momentum / mass
Velocity = 10 m/s / 0.145 kg
Velocity ≈ 68.97 m/s (rounded to two decimal places)

So, the velocity of the second ball just after it is struck is approximately 68.97 m/s.

its impulse it starts with is

mv=1.45 N. You know the mass of the ball, calculate its velocity. then, calculate its KEnergy.

Now you have two equations:
momentum
mv=mv1+mv2 unknowns v2, v1

energy
1/2 mv^2=1/2 mv1^2 + 1/2 mv2^2

so in equation 1, solve for v1 in terms of v2. then substitute all that for v1 in the second equation, and solve for v2 in the quadratic.