# A ball with a mass of #4# #kg# moving at #7# #ms^(-1)# hits a still ball with a mass of #7# #kg#. If the first ball stops moving, how fast is the second ball moving?

Before the collision, the first ball has all the momentum (since the second is stationary).

Momentum is conserved. After the collision the second ball has all the momentum.

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To find the velocity of the second ball after the collision, 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. The momentum ( p ) of an object is given by the product of its mass ( m ) and velocity ( v ), i.e., ( p = m \times v ).

Before the collision: ( p_{\text{initial}} = m_1 \times v_1 + m_2 \times v_2 )

After the collision: ( p_{\text{final}} = 0 \times v_1' + (m_1 + m_2) \times v_2' )

Since momentum is conserved, ( p_{\text{initial}} = p_{\text{final}} ), thus: ( m_1 \times v_1 + m_2 \times v_2 = (m_1 + m_2) \times v_2' )

Substituting the given values: ( (4 , \text{kg} \times 7 , \text{m/s}) + (7 , \text{kg} \times 0 , \text{m/s}) = (4 , \text{kg} + 7 , \text{kg}) \times v_2' ) ( 28 , \text{kg m/s} = 11 , \text{kg} \times v_2' ) ( v_2' = \frac{28 , \text{kg m/s}}{11 , \text{kg}} ) ( v_2' \approx 2.55 , \text{m/s} )

Therefore, the second ball is moving at approximately ( 2.55 , \text{m/s} ) after the collision.

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