Question

1. a toy rocket is launched from the ground by its blasting mechanism. the rocket had the ability to hit a height of 58 m while still travelling at 8.5 m/s. if the rocket lost 45% of its energy during its ascent, what was the initial velocity of the rocket? (4 marks-t) 3. a 52kg frozen block of ice at -16°c is heated to liquid water at 21°c a) draw the heat curve for this scenario (3 marks-c) b) how much heat energy was absorbed by the water during this process? (3 marks-a) ( c_{\text{ice}} = 2100 , j/kg^{circ}c ) ( c_{\text{water}} = 4186 , j/kg^{circ}c ) ( c_{\text{wain}} = 2.0 \times 10^{3} , j/kg^{circ}c ) ( l_{n} = 3.33 \times 10^{3} , j/kg ) ( l_{v} = 2.26 \times 10^{6} , j/kg ) ( q = mcdelta t ) ( q_{1} = 52 , kg cdot 2100 , j/c cdot (0 - (-16)) = 1747200 , j ) ( q_{2} = )

Explanation:

Step1: Calculate the mechanical energy at the height

The mechanical energy at height \(h = 58m\) with velocity \(v=8.5m/s\) is the sum of potential energy \(E_p=mgh\) and kinetic energy \(E_k=\frac{1}{2}mv^{2}\). Let the mass of the rocket be \(m\). Then \(E = mgh+\frac{1}{2}mv^{2}\), where \(g = 9.8m/s^{2}\), \(h = 58m\), \(v = 8.5m/s\). So \(E=m(9.8\times58+\frac{1}{2}\times8.5^{2})=m(568.4 + 36.125)=m\times604.525\)

Step2: Relate the initial energy to the final energy

If the rocket lost \(45\%\) of its energy during ascent, then the final energy \(E\) is \((1 - 0.45)E_{0}\), where \(E_{0}\) is the initial energy. So \(E_{0}=\frac{E}{0.55}\)

Since \(E_{0}=\frac{1}{2}mv_{0}^{2}\) (initial energy is kinetic, assuming launched from rest in terms of potential - ground level \(h = 0\)), and \(E=m\times604.525\)

We have \(\frac{1}{2}mv_{0}^{2}=\frac{m\times604.525}{0.55}\)

Cancel out \(m\) (mass cancels as it is non - zero)

\(\frac{1}{2}v_{0}^{2}=\frac{604.525}{0.55}\)

\(\frac{1}{2}v_{0}^{2}= 1099.136\)

\(v_{0}^{2}=2\times1099.136 = 2198.272\)

\(v_{0}=\sqrt{2198.272}\approx46.9m/s\)

Answer:

The initial velocity of the rocket is approximately \(46.9m/s\)