A ball with a mass of #2kg# moving at #1 m/s# hits a still ball with a mass of #3 kg#. If the first ball stops moving, how fast is the second ball moving?

Question

A ball with a mass of #2kg# moving at #1 m/s# hits a still ball with a mass of #3 kg#.  If the first ball stops moving, how fast is the second ball moving?

Charlotte Aaron · Accepted Answer

The answer is #=2/3ms^(-1)#

Here it's an elastic collision with no loss of kinetic energy.

We have the conservation of momentum.

#m_1u_1+m_2u_2=m_1v_1+m_2v_2#

#m_1=2 kg#

#u_1=1 ms^(-1)#

#m_2=3 kg#

#u_2=0#

#v_1=0#

#v_2=?#

#2*1+3*0=0+3v_2#

#v_2=2/3 ms^(-1)#

Xavier Adame · Answer

undefined We can use the principle of conservation of momentum to solve this problem. According to this principle, the total momentum of an isolated system remains constant if no external forces act on it.

The momentum of an object is given by the product of its mass and velocity (p = mv).

Initially, the momentum of the system is the sum of the momenta of both balls:

$ p_{\text{initial}} = m_1v_1 + m_2v_2 $

Where:
- $ m_1 = 2 $ kg (mass of the first ball)
- $ v_1 = 1 $ m/s (initial velocity of the first ball)
- $ m_2 = 3 $ kg (mass of the second ball)
- $ v_2 $ is the velocity of the second ball, which we need to find

After the collision, the first ball stops moving, so its final velocity ($ v_{1f} $) is 0 m/s. Therefore, the momentum of the system becomes:

$ p_{\text{final}} = 0 + m_2v_{2f} $

Since momentum is conserved, we can set the initial momentum equal to the final momentum:

$ m_1v_1 + m_2v_2 = 0 + m_2v_{2f} $

Substituting the given values:

$ 2 \times 1 + 3 \times 0 = 0 + 3 \times v_{2f} $

Solving for $ v_{2f} $:

$ 2 = 3v_{2f} $

$ v_{2f} = \frac{2}{3} $ m/s

Therefore, the second ball is moving at $ \frac{2}{3} $ m/s after the collision.