Question

You throw a metal block of mass \(0.25\;\;{\rm{kg}}\)into the air, and it leaves your hand at time\(t = 0\)at location\(\langle 0,2,0\rangle \;{\rm{m}}\)with velocity\(\langle 3,4,0\rangle \;{\rm{m}}/{\rm{s}}\). At this low velocity air resistance is negligible. Using the iterative method shown in Section\(2.4\)with a time step of\(0.05\;\;{\rm{s}}\),calculate step by step the position and velocity of the block at\(t = 0.05\;\;{\rm{s}},t = 0.10\;\;{\rm{s}}\), and\(t = 0.15\;\;{\rm{s}}\)

Short answer

The position and velocity with respect to time are shown

\(t = 0.05\;{\rm{s}}\)_,\({\vec r_f} = \langle 0.15,2.176,0\rangle {\rm{m}}\)\({\vec v_f} = \langle 3,3.512,0\rangle {\rm{m}}/{\rm{s}}\)_.

\(t = 0.10\;{\rm{s}}\)_,\({\vec r_f} = \langle 0.3,2.327,0\rangle {\rm{m}}\)_,\({\vec v_f} = \langle 3,3.024,0\rangle {\rm{m}}/{\rm{s}}\)_.

\(t = 0.15\;{\rm{s}}\)_, \({\vec r_f} = \langle 0.45,2.454,0\rangle {\rm{m}}\)\({\vec v_f} = \langle 3,2.536,0\rangle {\rm{m}}/{\rm{s}}\)_.

Step-by-step solution

Given

A metal block of mass \(0.25\;\;{\rm{kg}}\)into the air, and it leaves your hand at time\(t = 0\)at location\(\langle 0,2,0\rangle \;{\rm{m}}\)with velocity\(\langle 3,4,0\rangle \;{\rm{m}}/{\rm{s}}\)

Define the Iterative Method

The Iterative Method is a mathematical way of solving a problemby repeatedly using the value of previous solution in the next solution it starts from one assume value.

Step 3: The position and velocity of the block at \(\Delta t = 0.05\;{\rm{s }}\)

There is no interaction between your hand and the block after it leave your hand, and since we ignore air resistance, the only force that acts on the block is the gravitational force, so

\(\begin{aligned}{c}{{\vec F}_{{\rm{net }}}} &= {{\vec F}_g}\\ &= m\vec g\\ &= (0.25)\langle 0, - 9.8,0\rangle \\ = \langle 0, - 2.45,0\rangle {\rm{N}}\end{aligned}\)

The future momentum the will had at the end of this time step is

\(\begin{aligned}{c}{{\vec p}_f} &= {{\vec p}_i} + {{\vec F}_{{\rm{net }}}}\Delta t\\ &= \langle 0.75,1,0\rangle + \langle 0, - 2.45,0\rangle (0.05)\\ &= \langle 0.75,0.878,0\rangle {\rm{kg}} \cdot {\rm{m}}/{\rm{s}}\end{aligned}\)

\(\begin{aligned}{c}{{\vec v}_f} &= \frac{{{{\vec p}_f}}}{m}\\ &= \frac{{\langle 0.75,0.878,0\rangle }}{{0.25}}\end{aligned}\)

\(\begin{aligned}{c}{{\vec v}_f} &= \frac{{{{\vec p}_f}}}{m}\\ &= \frac{{\langle 0.75,0.878,0\rangle }}{{0.25}}\\ &= \langle 3,3.512,0\rangle {\rm{m}}/{\rm{s}}\end{aligned}\)

The velocity at time\(t = 5\;{\rm{s}}\)is\(\langle 3,3.512,0\rangle \;{\rm{m}}/{\rm{s}}\).