Answer to Question #156178 in Math for Tony

Question

Three vectors a = (3i 4j), b = (-4i + 3j), c = (-3i + 4j) lie on a horizontal plane, pie(ll).
(i) Determine which of the vectors b or a is perpendicular to a.
Two smooth spheres, S and T, each of mass m lie on the plane ll. Sphere S is projected along the plane towards sphere T with velocity 5x(i + 2j) so that it collides obliquely with T. At the instant of collision, the line of impact is parallel to the vector a. Given that the coefficient of restitution between the two spheres is 1/2,
Show that
(ii) the component of the initial velocity of S parallel to a is 11x.
(iii) the component of the final velocity os S parallel to a is (11/4)x.
(iv) the magnitude of the impulse exerted by sphere T on sphere S is (33/4)mx.

Answer 1

(i) Two vectors are said to be perpendicular if their dot product is zero.

$a \cdot b=(3i+4j) \cdot(-4i+3j)=-12+12=0\\ a \cdot c=(3i+4j) \cdot(-3i+4j)=-9+16=7$

Since $a \cdot b=0$, this implies that $b$ is the vector perpendicular to $a$ \\

(ii) The initial velocity $u$ of $S$ is $(5xi+10xj)$.

The component of the initial velocity of $S$ parallel to $a$ is $\frac{\overrightarrow{u}\cdot a}{|a|}$

$\frac{(5xi+10xj).(3i+4j)}{\sqrt{3^2+4^2}}=\frac{15x+40x}{5}=\frac{55x}{5}=11x$

(iii) We need to compute the final velocity of $S$.

$m_1u_1+m_2u_2=m_1v_1+m_2v_2\\ m(5xi+10xj+0)=m(v_1+v_2)\\ 5xi+10xj=v_1+v_2................................(1)\\$

The coefficient of restitution $e$ =$-\frac{v_2-v_1}{u_2-u_1}$

$\frac{1}{2}=-\frac{v_2-v_1}{0-(5xi+10xj)}\\ \frac{1}{2}(5xi+10xj)=v_2-v_1...........................(2)$

Add (1) and (2) together

$2v_2=\frac{15x}{2}i+15xj\\ v_2=\frac{15x}{4}i+\frac{15x}{2}j$

Substitute this into (1)

$v_1=5xi+10xj-(\frac{15x}{4}i+\frac{15x}{2}j)\\ v_1=\frac{5x}{4}i+\frac{5x}{2}j$

The final velocity $v_1$ of $S$ is $\frac{5x}{4}i+\frac{5x}{2}j$

$\frac{v_1 \cdot a}{|a|}=\frac{(\frac{5x}{4}i+\frac{5x}{2}j)\cdot (3i+4j)}{5}=\frac{\frac{55x}{4}}{5}=\frac{11x}{4}$

(iv) The impulse exerted by $T$ on $S$ is $m(v_2-u_2)$

$=m(\frac{15x}{4}i+\frac{15x}{2}j-0)\\ =m(\frac{15x}{4}i+\frac{15x}{2})$

Magnitude=$\sqrt{m^2[(\frac{15x}{4})^2+(\frac{15x}{2})^2]}$

$=m\sqrt{\frac{1125x^2}{16}}=\frac{15\sqrt{5}mx}{4}$