**Conservation of Momentum**

Object A with mass \(m_A\) is moving with a velocity \(v_A\). Object B with a mass \(m_B\) moves with a velocity of \(v_B\). The two objects collide and after collision move with velocities \(v'_A\) and \(v'_B\) respectively. What is the momentum of the objects before and after they collide?

Given the momentum of an object as \(p=mv\), substitute the values of the mass and velocity of each object before and after collision. Thus,

Before collision: \(p_A=m_Av_A\) and \(p_B=m_Bv_B\).

After collision: \(p_A=m_Av'_A\) and \(p_B=m_Bv'_B\).

Combining the equations of each object before and after collision will give the change in momentum of the objects.

\(\Delta p_A=m_Av_A-m_Av'_A\) (1)

\(\Delta p_B=m_Bv_B-m_Bv'_B\) (2)

What causes the change of momentum of the objects? - Force. The force exerted by object B as it collides with A changes A's momentum and at the same time the force exerted by A to B causes the change in momentum of B. This can be expressed mathematically using the equation for impulse, \(F\Delta t=\Delta p\).

\(F_{B-A}\Delta t=\Delta p_A=m_Av_A-m_Av'_A\) (3)

\(F_{A-B}\Delta t=\Delta p_B=m_Bv_B-m_Bv'_B\) (4)

Newton's third law states that two objects interacting with each other applies equal force but of opposite direction, thus \(F_{B-A}=-F_{A-B}\), the negative sign indicates the opposite direction of the force. Hence, the impulse can be written as

\(F_{B-A}\Delta t=-F_{A-B}\Delta t\) (5)

Substituting (3) and (4) to (5) gives the equation \(m_Av_A-m_Av'_A=-(m_Bv_B-m_Bv'_B)\). Simplifying,

\(m_Av_A+m_Bv_B=m_Av'_A+m_Bv'_B\) (6)

The left side of the equation indicates the total momentum before collision and the right side as the total momentum of the objects after collision. Equation (6) describes that the total momentum before collision is equal to the total momentum after the collision. This law is called **conservation of momentum. **Momentum is conserved if there is no other external forces acting on the objects aside from the force they exert to each other.

**Example 1.**

Object A moving at \(4\;m/s\) with a mass of \(2 \;kg\)** **interacts with another object B with mass \(6\;kg\) moving at \(8\;m/s\). What is the final velocity of object B if A is moving at \(10\;m/s\) after the collision?

Given:

\(m_A=2\;kg \quad\quad\quad\quad m_B=6\;kg\\ v_A=4\;m/s \quad\quad\quad\quad v_B=8\;m/s\\ v'_A=10\;m/s \)

Solution:

\(m_Av_A+m_Bv_B=m_Av'_A+m_Bv'_B\)

\((2\;kg)(4\;m/s)+(6\;kg)(8\;m/s)=(2\;kg)(10\;m/s)+(6\;kg)(v'_B)\)

\((8\;kg\cdot m/s)+(48\;kg\cdot m/s)=(20\;kg\cdot m/s)+(6\;kg)(v'_B)\)

\(56\;kg\cdot m/s-20\;kg\cdot m/s=(6\;kg)(v'_B)\)

\(36\;kg\cdot m/s=(6\;kg)(v'_B)\)

\(v'_B=\frac {36\;kg\cdot m/s}{6\;kg}=6\;m/s\)

thus, object B is moving at \(6\;m/s\) after collision.

**Example 2.**

A golf club with a mass of \(0.33\;kg\) is about to hit a golf ball with a mass of \(0.045\;kg\). From rest, the golf ball move at a speed of \(61.2\;m/s\) after it was hit by the club. Solve for the velocity of the club as it hit the ball. Assume that the club is at rest after hitting the ball.

Given:

\(m_c=0.33\;kg \quad\quad\quad\quad m_b=0.045\;kg\\ v'_A=0\;m/s\quad\quad\quad\quad\quad v_b=0\;m/s\\ \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad v'_b=61.2\;m/s \)

Solution:

\(m_cv_c+m_bv_b=m_cv'_c+m_bv'_b\)

\((0.33\;kg)(v_c)+(0.045\;kg)(0\;m/s)=(0.33\;kg)(0\;m/s)+(0.045\;kg)(61.2\;m/s)\)

\((0.33\;kg)(v_c)=2.75\;kg\cdot m/s\)

\(v_c=\frac{2.75\;kg\cdot m/s}{0.33\;kg}=8.33\;m/s\)

Therefore, the initial velocity of the golf club is \(8.33\;m/s\).