Question

A frictionless steel ball of radius $2cm$, moving on a horizontal plane with a velocity of $5cm/s$, collides head-on with another stationary steel ball of radius $3cm$. The velocities of two bodies after collision will respectively be (in cm/s) ($e=1$)-

• A. $2.7,2.3$
• B. $-2.7,2.3$
• C. $2.7,-2.3$
• D. $-2.7,-2.3$

Answer 1

The correct option is B $-2.7,2.3$

As both balls of steel so same density and so their mass will be proportional to their volume and volume is proportional to cube of radius so their mass $m_1$ and $m_2$ will be in ratio of $2^3$ :$3^3$ i.e. $8:27$

so suppose $m_1=8m$ and $m_2=27m$ then use equation

$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$ and Newton's law of collision as $\frac{u_1-u_2}{v_2-v_1}$=$e$ ............................1

given $e=1$ and $u_1=5cm/s$ $u_2=0$

From the equation-1 we get $v_2=v_1+5$ put it in the momentum conservation equation we will get $v_1=-2.7cm/s$ and $v_2=2.3cm/s$

Note- Negative sign shows that the lighter mass bounced means went back with less speed .