Question

A 0.600-kg ball traveling at 4.00 m/s to the right collides with a 1.00-kg ball traveling at 5.00 m/s to the left. After the collision, the lighter ball is traveling 7.25 m/s to the left. What is the velocity of the heavier ball after the collision?​

Answer 1

Step-by-step explanation:

The velocity of each ball after the collision can be determined using the law of conservation of momentum. The law of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces are acting on the system. In this case, the two balls form an isolated system before and after the collision, so their total momentum must be conserved.

Let's call the velocity of the heavier ball after the collision "v". The total momentum before the collision is given by:

p_i = (0.600 kg)(4.00 m/s) + (1.00 kg)(-5.00 m/s) = -2.00 kg m/s

The total momentum after the collision is given by:

p_f = (0.600 kg)(v) + (1.00 kg)(-7.25 m/s) = 0.600 kg v - 7.25 kg m/s

Since the total momentum is conserved, we can set the initial momentum equal to the final momentum and solve for v:

p_i = p_f

-2.00 kg m/s = 0.600 kg v - 7.25 kg m/s

Adding 7.25 kg m/s to both sides and dividing both sides by 0.600 kg, we get:

v = (2.00 kg m/s + 7.25 kg m/s) / 0.600 kg = 9.25 m/s

So, the velocity of the heavier ball after the collision is 9.25 m/s.

Answer 2

The velocity of the heavier ball after the collision is 3.05 m/s

Step-by-step explanation:

This is an example of partially elastic collision. In a partially elastic collision, momentum is conserved and the objects move at different speeds. A portion of the initial kinetic energy is still retained by the colliding objects after the impact.

The conservation of momentum states that the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

The formula for momentum is

`p=mv`

We can conclude that

`m_1v_(1i) +m_2v_(2i)=m_1v_(1f) +m_2v_(2f)`

Let object 1 be the 0.600 kg ball.

Let object 2 be the 1.00 kg ball.

We are given

`m_1=0.600\\v_(1i) =4.00\\m_2=1.00\\v_(2i) =5.00\\v_(1f) =7.25`

Inserting our given values into the equation gives us

`(0.600*4.00)+(1.00*5.00)=(0.600*7.25)+\left(1.00*v_(2f)\right)`

Lets solve for
`v_(2f)`.

Evaluate each set of parenthesis.

`2.4+(1.00*5.00)=(0.600*7.25)+\left(1.00*v_(2f)\right)`

`2.4+5=(0.600*7.25)+\left(1.00*v_(2f)\right)`

`2.4+5=4.35+\left(1.00*v_(2f)\right)`

Anything multiplied by 1 stays the same.

`2.4+5=4.35+v_(2f)`

Add 2.4 and 5.

`7.4=4.35+v_(2f)`

Subtract 4.35 from both sides of the equation.

`3.05=v_(2f)`