Question

a 3kg ball and 4kg ball having speed 7m/s and 5m/s approach one another. Find their speeds after collision if the coefficient of restitution is 0.75.

Answer 1

mass of first ball, $m_1$= 3kg

speed of first ball, $u_1$ = 7m/s

mass of 2nd ball, $m_2$ = 4kg

speed of 2nd ball, $u_2$ = -5m/s [balls are approaching]

Also given, coefficient of restitution, e = 0.75 = $\frac{v_2-v_1}{u_1-u_2}$

or, 0.75 = $\frac{v_2-v_1}{7+5}$

or, 9 = $v_2-v_1$.......(1)

now, applying law of conservation of linear momentum,

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

or, 3 × 7 + 4 × (-5) = $3v_1+4v_2$

or, $3v_1+4v_2=1$......(2)

from equations (1) and (2),

$3v_1+4(9+v_1)=1$

$7v_1=-35\implies v_1=-5$m/s

and $v_2=9-5=4$ m/s

speed of first ball is -5m/s and speed of 2nd ball is 4m/s