Question

An object with mass 3 kg moving with an initial velocity of 500i m/s collided with and sticks to an object of mass 2 kg with an initial mass of -3j m/s. Find the final velocity of the composite object

Answer 1

Explanations & Calculations

• Here the two objects were moving along two directions orthogonal to each other. (along the positive direction of the x-axis & the negative y-axis)
• The initial momentum of the system along x is

$\qquad\qquad \begin{aligned} \small P_{1x}&=\small 3kg\times 500\it{i}\,ms^{-1}\\ &=\small 1500\it{i}\,\,kgms^{-1} \end{aligned}$

• That along the y-direction is

$\qquad\qquad \begin{aligned} \small P_{1y}&=\small 2\,kg\times (-3\it{j})ms^{-1}\\ &=\small -6\it{j}\,kgms^{-1} \end{aligned}$

• Final mass of the system is $\small 5\,kg$.
• Momentum is conserved during this collision hence its final horizontal & vertical momenta should be correspondingly equal to the initial momenta.
• Therefore,

$\qquad\qquad \begin{aligned} \small P_{2x}&=\small P_{1x}\\ \small 5kg\times \vec{V_x}&=\small 1500\it{i}\\ \small \vec{V_x}&=\small 300\it{i}\,ms^{-1}\\\\ \small P_{2y}&=\small P_{1y}\\ \small 5kg\times \vec{V_y}&=\small -6\it{j}\\ \small \vec{V_y} &=\small -1.2\it{j}\,ms^{-1} \end{aligned}$

• Then the final velocity of the combined object in vector form is

$\qquad\qquad \begin{aligned} \small \vec{V}&=\small 300\it{i}-1.2\it{j} \end{aligned}$

• Then the magnitude & the direction of it are

$\qquad\qquad \begin{aligned} \small |\vec{V}|&=\small \sqrt{300^2+1.2^2}=\bold{300.0024\,ms^{-1}}\\\\ \small \theta&=\small \tan^{-1}\bigg[\frac{-1.2}{300}\bigg]=\bold{-0.229^0} \end{aligned}$

• Then $\small \vec{V}=300.00\,ms^{-1}\,[\,S\,0.229^0\,E\,]$