# A) A child on a swing is pulled back by daddy and released from rest at a height of 0.93 m above...

## Question:

A) A child on a swing is pulled back by daddy and released from rest at a height of 0.93 m above the swing's lowest point. Find the child's speed as the swing passes through its lowest point. m/s

B) Two hockey pucks of equal mass approach each other head-on with equal speeds of 8.4 m/s. They collide and one puck leaves the collision with a speed of 6.4 m/s. What is the speed of the other puck after the collision? m/s

## Conservation of energy and momentum

For the first problem, we will apply the conservation of energy, which states that the initial energy of the system is equal to its final energy {eq}\rm\ E_{initial} = \rm\ E_{final}{/eq}

For the second problem, we apply the law of conservation of momentum, which states that without external forces, the momentum is conserved {eq}\vec P_i = \vec P_f{/eq}

## Answer and Explanation:

Part A:

Given:

• Height {eq}y = 0.93 \rm\ m{/eq}

We apply the conservation of energy to find the velocity of the child. Before the child is released, their energy is entirely potential energy. At the lowest point of their swing, all their energy will have been converted to kinetic energy

\begin{align} PE_i + KE_i &=PE_f + KE_f\\ mgy &= \frac{1}{2}mv^2\\ v &= \sqrt{2gy}\\ v &= \sqrt{2(9.8 \rm\ m/s^2 \cdot 0.93 \rm\ m)}\\ v &= \boxed{4.27 \rm\ m/s} \end{align}

Part B:

Given:

• Initial velocity of each puck {eq}v_1 = v_2 = 8.4 \rm\ m/s{/eq}
• Final velocity of puck 1 {eq}v_1' = 6.4 \rm\ m/s{/eq}

Apply the law of conservation of momentum to find the velocity of the other puck. Note that they have the same mass

\begin{align} m_2v_1 + m_2v_2 &= m_1v_1' + m_2v_2'\\ v_1 + v_2 &= v_1' + v_2'\\ v_2' &= v_1 + v_2 - v_1'\\ v_2' &= 8.4 \rm\ m/s + 8.4 \rm\ m/s - 6.4 \rm\ m/s\\ v_2' &= \boxed{10.4 \rm\ m/s} \end{align}