Question

Two balls of 0.4 kg and 0.2 kg mass with a velocity of 2 m/s and 4 m/s, respectively, approach each other to execute a perfectly elastic collision. Find the velocity of eacha aftercollision.

Answer 1

 From the law of conservation of momentum, we have:

$$m_1u_1+m_2u_2=m_1v_1+m_2v_2. (1)$$

Since collision is perfectly elastic, kinetic energy is conserved and we can write:

$$\dfrac{1}{2}m_1u_1^2+\dfrac{1}{2}m_2u_2^2=\dfrac{1}{2}m_1v_1^2+\dfrac{1}{2}m_2v_2^2. (2)$$

Let’s rearrange equations (1) and (2):

$$m_1(u_1-v_1)=m_2(v_2-u_2), (3)$$ $$m_1(u_1^2-v_1^2)=m_2(v_2^2-u_2^2). (4)$$

Let’s divide equation (4) by equation (3):

$$\dfrac{(u_1-v_1)(u_1+v_1)}{u_1-v_1}=\dfrac{(v_2-u_2)(v_2+u_2)}{v_2-u_2},$$ $$u_1+v_1=u_2+v_2. (5)$$

Let's express $v_2$ from the equation (5) in terms of $u_1$, $u_2$ and $v_1$:

$$v_2=u_1-u_2+v_1. (6)$$

Let’s substitute equation (6) into equation (3). After simplification, we get:

$$(m_1-m_2)u_1+2m_2u_2=(m_1+m_2)v_1.$$

From this equation we can find velocity of first ball after collision, $v_1$:

$$v_1=\dfrac{(m_1-m_2)}{(m_1+m_2)}u_1+\dfrac{2m_2}{(m_1+m_2)}u_2,$$ $$v_1=\dfrac{(0.4\ kg-0.2\ kg)}{(0.4\ kg+0.2\ kg)}\cdot2\ \dfrac{m}{s}+\dfrac{2\cdot0.2\ kg}{(0.4\ kg+0.2\ kg)}\cdot4\ \dfrac{m}{s},$$ $$v_1=3.3\ \dfrac{m}{s}.$$

Substituting $v_1$into the equation (6) we can find velocity of second ball after collision, $v_2$:

$$v_2=\dfrac{2m_1}{(m_1+m_2)}u_1+\dfrac{(m_2-m_1)}{(m_1+m_2)}u_2,$$ $$v_2=\dfrac{2\cdot0.4\ kg}{(0.4\ kg+0.2\ kg)}\cdot2\ \dfrac{m}{s}+\dfrac{(0.2\ kg-0.4\ kg)}{(0.4\ kg+0.2\ kg)}\cdot4\ \dfrac{m}{s},$$ $$v_2=1.33\ \dfrac{m}{s}.$$