Question

Consider a glancing collision between two identical spheres of 0.048 kg, with one of the spheres initially at rest. Initially the incoming projectile has a velocity of 1.50 m/s to the right and after the collision we observe that this sphere leaves the collision region with a velocity of 0.40 m/s at an angle of 33° to its initial direction. (Assume that the +x-axis is to the right and the +y-axis is up along the page. Angles are measured counterclockwise from the +x-axis. Do not assume this collision is elastic.) (a) Find the x-component of the velocity of the second sphere. (b) Find the y-component of the velocity of the second sphere.

Answer 1

Mass of both spheres, m= 0.048kg

Assuming the given directions: +x-axis to the right and +y-axis is up along the page

Then vector notation of Initial velocity of 1st sphere, $\vec{u}_1= 1.50 \hat{i}$

Vector notation of Initial velocity of 2nd sphere, $\vec{u}_2= 0$

Vector notation of final velocity of 1st sphere, $\vec{v}_1= 0.40cos33^\circ\hat{i}+0.40sin33^\circ\hat{j}$

   $\vec{v}_1= 0.335\hat{i}+0.217\hat{j}$

Let vector notation of final velocity of 2nd sphere be $\vec{v}_2$

Then using Conservation of Momentum Principle,

$\vec{p}_{initial}= \vec{p}_{final}$

$m_1\vec{u}_1+m_2\vec{u}_2=m_1\vec{v}_1+m_2\vec{v}_2$

   $m(1.50\hat{i}+0)=m(0.335\hat{i}+0.217\hat{j}+\vec{v}_2)$

$\vec{v}_2= 1.50\hat{i}-(0.335\hat{i}+0.217\hat{j})$

$\vec{v}_2= 1.165\hat{i}-0.217\hat{j}$ (ANS)

a). Then x-component of the velocity of 2nd sphere will be

(ANS)

b). Then y-component of the velocity of 2nd sphere will be

  (ANS) (Negative sign shows velocity towards negative y-axis)