A ball with a mass of #4 kg# moving at #6 m/s# hits a still ball with a mass of #8 kg#. If the first ball stops moving, how fast is the second ball moving?

Question

A ball with a mass of #4 kg# moving at #6 m/s# hits a still ball with a mass of #8 kg#.  If the first ball stops moving, how fast is the second ball moving?

Elena Albarran · Accepted Answer

#3m/s#

Conservation of momentum states that there must be the same amount of momentum before and after a reaction. Momentum is given by the equation

#p=mv#

where #p# is momentum, #v# is velocity and #m# is mass.

Before the reaction, momentum is

#p=mv# #p=4kg*6m/s=24(kgm)/s#

After the reaction, there must the same momentum of #24(kgm)/s#. Since the first ball has stopped moving, it has transferred all of its momentum to the second #8kg# ball.

#p=mv# #24(kgm)/s=8kg*xm/s#

Rearranging, we find that

#x=24/8m/s=3m/s#

Emily Abbate · Answer

undefined To find the velocity of the second ball after the collision, you can use the principle of conservation of momentum. The total momentum before the collision equals the total momentum after the collision. Therefore, you can calculate it as follows:

$ m_1 \cdot v_1 + m_2 \cdot v_2 = (m_1 + m_2) \cdot v_f $

Where:
$ m_1 = 4 \, \text{kg} $ (mass of the first ball)
$ v_1 = 6 \, \text{m/s} $ (initial velocity of the first ball)
$ m_2 = 8 \, \text{kg} $ (mass of the second ball)
$ v_2 $ is the initial velocity of the second ball (which is zero)
$ v_f $ is the final velocity of both balls after the collision.

By substituting the given values into the equation, you can solve for $ v_f $, which will give you the final velocity of both balls after the collision.

Amelia Adams · Answer

undefined To solve this problem, we can use the principle of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision in an isolated system.

The formula for momentum is $ momentum = mass \times velocity $.

Before the collision, the momentum of the system is the sum of the momentums of both balls:

$ momentum_{\text{before}} = (4 \, \text{kg} \times 6 \, \text{m/s}) + (8 \, \text{kg} \times 0 \, \text{m/s}) $

After the collision, since the first ball stops moving, its final velocity is 0 m/s, and the second ball moves with a final velocity $ v_2 $ (which we need to find).

So, the momentum after the collision is:

$ momentum_{\text{after}} = (0 \, \text{kg} \times 0 \, \text{m/s}) + (8 \, \text{kg} \times v_2) $

According to the principle of conservation of momentum:

$ momentum_{\text{before}} = momentum_{\text{after}} $

Solving for $ v_2 $:

$ (4 \, \text{kg} \times 6 \, \text{m/s}) = (8 \, \text{kg} \times v_2) $

$ 24 \, \text{kg m/s} = 8 \, \text{kg} \times v_2 $

$ v_2 = \frac{24 \, \text{kg m/s}}{8 \, \text{kg}} $

$ v_2 = 3 \, \text{m/s} $

So, the second ball is moving at a speed of $ 3 \, \text{m/s} $ after the collision.